Q.A cricket ball of mass 150 g has an initial velocity $\mathbf{u} = (3\hat{i} + 4\hat{j})$ m s$^{-1}$ and a final velocity $\mathbf{v} = -(3\hat{i} + 4\hat{j})$ m s$^{-1}$ after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s$^{1}$)

Showing the 12 most recent of 28 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The figure shows the position-time graph of a particle of mass 2 kg [graph: $x(m)$ vs $t(s)$; position increases linearly from the origin $(0,0)$ to point A at $(t=2\text{ s}, x=4\text{ m})$, then remains constant at $x=4$ m for $t>2$ s]. The impulse of the particle at $t=0$ s is (A) $2\ kg\,ms^{-1}$ (B) $4\ kg\,ms^{-1}$ (C) $1\ kg\,ms^{-1}$ (D) $3\ kg\,ms^{-1}$

   ›Reveal solutionSolution

   The particle is jolted from rest into a constant velocity of 2 m/s at $t=0$; impulse equals the resulting change in momentum, $4$ kg·m/s.

   Concept and Intuition

   Impulse is defined as the change in linear momentum, $J=\Delta p=m(v_f-v_i)$. A position-time graph's slope directly gives velocity, so reading off the slope in each segment of the graph tells us the velocities before and after any sudden change (here, at $t=0$).

   Step-by-Step Solution

   1. From the graph, for $0\le t\le2$ s, position increases linearly from $x=0$ to $x=4$ m, so the velocity in this interval is $v=\dfrac{\Delta x}{\Delta t}=\dfrac{4}{2}=2$ m/s.
   2. The phrase "impulse of the particle at $t=0$ s" refers to the sudden change imparted right at the start — the particle goes from being at rest (implicit, since it starts exactly at the origin with no prior motion shown) to moving at $2$ m/s.
   3. So $\Delta v = v_f - v_i = 2 - 0 = 2$ m/s.
   4. Impulse $=m\Delta v = 2\ \text{kg}\times 2\ \text{m/s} = 4$ kg·m/s.

   Common Mistakes

   • Confusing the impulse over the whole 0–2 s interval (which would be a force$\times$time calculation, not needed here) with the instantaneous impulse at $t=0$ that sets the motion going.
   • Forgetting to multiply by the mass (using $\Delta v$ alone as the answer).

   ✓Final answer

   The correct option is (B) — $4\ kg\,ms^{-1}$.

   ANSWER: B

2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A ball of mass M moving with a velocity of $4\ m\,s^{-1}$ collides with another ball of mass 2M moving in the same direction. If the coefficient of restitution between the two balls is 0.35 and the velocity of the heavier ball after collision is $2.9\ m\,s^{-1}$, then the initial velocity of the heavier ball is (A) $3\ m\,s^{-1}$ (B) $4\ m\,s^{-1}$ (C) $6\ m\,s^{-1}$ (D) $2\ m\,s^{-1}$

   ›Reveal solutionSolution

   This is a one-dimensional collision with a known coefficient of restitution and one unknown initial velocity plus one unknown final velocity — solve the two simultaneous equations (momentum conservation + Newton's restitution law) together. Answer: (D).

   Concept and Intuition

   A collision between two bodies always obeys two independent physical laws: (1) total momentum is conserved (no external horizontal force acts during the brief collision), and (2) the coefficient of restitution $e$ relates the relative velocity of separation after collision to the relative velocity of approach before collision, $e=\dfrac{\text{velocity of separation}}{\text{velocity of approach}}$. Here we're told the lighter ball's initial speed (4 m/s) and the heavier ball's final speed (2.9 m/s), but not the lighter ball's final speed nor the heavier ball's initial speed — that's two unknowns, which is exactly what our two equations can solve for.

   Step-by-Step Solution

   1. Set up variables. Let $u_1=4\ m/s$ (initial velocity of mass $M$), $u_2=$ initial velocity of mass $2M$ (unknown, to find), $v_1=$ final velocity of $M$ (unknown), $v_2=2.9\ m/s$ (given final velocity of $2M$). All velocities are along the same line (same direction, as stated).
   2. Momentum conservation:

   $M u_1 + 2M u_2 = M v_1 + 2M v_2$

   $4 + 2u_2 = v_1 + 5.8 \quad\Rightarrow\quad v_1 = 2u_2 - 1.8. \qquad(i)$

   3. Restitution equation (separation speed $=e\times$ approach speed):

   $e=\frac{v_2-v_1}{u_1-u_2}\quad\Rightarrow\quad 0.35 = \frac{2.9-v_1}{4-u_2}.\qquad(ii)$

   4. Substitute (i) into (ii):

   $0.35(4-u_2)=2.9-(2u_2-1.8)=4.7-2u_2$

   $1.4-0.35u_2=4.7-2u_2$

   $1.65u_2=3.3 \quad\Rightarrow\quad u_2=2\ m/s.$

   5. Check: $v_1=2(2)-1.8=2.2\ m/s$. Restitution check: $(2.9-2.2)/(4-2)=0.35$ ✓. Momentum check: $4(1)+2(2)=8$ and $2.2(1)+2(2.9)=8$ ✓ (in units of $M$).

   Common Mistakes

   • Assuming the heavier ball was initially at rest (a common simplification in easier problems) — here it's explicitly moving, and that velocity is exactly what's asked.
   • Getting the sign convention wrong in the restitution formula (approach speed must be positive since the lighter ball catches up to the slower heavier ball, $u_1>u_2$, consistent with $4>2$).
   • Trying to solve with momentum alone — one equation cannot fix two unknowns; the restitution relation is essential here.

   ✓Final answer

   The correct option is (D) — $2\ m\,s^{-1}$.

   ANSWER: D

3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A stream of water flowing horizontally with a speed of $15\ m\,s^{-1}$ gushes out of a tube of cross-sectional area $10^{-2}\ m^2$ and hits a vertical wall nearby. Assuming water does not rebound, The force exerted on the wall by the impact of water is (A) 2.25 N (B) $2.25\times10^3$ N (C) 1.5 N (D) $1.5\times10^3$ N

   ›Reveal solutionSolution

   Force from a water jet hitting a wall (no rebound) equals the rate of momentum transfer, $\rho A v^2$; numerically this comes out to $2.25\times10^3$ N.

   Concept and Intuition

   When a stream of water hits a wall and does not bounce back, every bit of water arriving loses all its forward momentum. By Newton's second law in the impulse-momentum form, the force on the wall equals the rate at which momentum is destroyed, which is the mass flow rate times the velocity.

   Step-by-Step Solution

   1. Mass flow rate of water: $\dot{m} = \rho A v$, where $\rho = 1000\ kg/m^3$ (density of water), $A = 10^{-2}\ m^2$, $v = 15\ ms^{-1}$.
   2. $\dot{m} = 1000 \times 10^{-2} \times 15 = 150\ kg/s$.
   3. Since the water does not rebound, its momentum drops from $mv$ to $0$ upon impact, so the rate of momentum change (= force on water, and by Newton's third law, force on wall) is: $F = \dot{m} \times v = 150 \times 15 = 2250\ N$.
   4. $2250\ N = 2.25\times10^3\ N$.

   Common Mistakes

   • Forgetting to square the velocity implicitly (using $F = \rho A v$ instead of $\rho A v^2$) — a very common shortcut error.
   • Assuming the water rebounds elastically (doubling the force) — the problem explicitly states it does not rebound.

   ✓Final answer

   The correct option is (B) — $2.25\times10^3$ N.

   ANSWER: B

4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.A bullet fired from a gun is given a force $(600 - 2 \times 10^5\, t)$ newton, where 't' is in second. Force on the bullet becomes zero when it leaves the barrel. The average impulse imparted to the bullet is (A) 9 Ns (B) zero (C) 0.9 Ns (D) 1.8 Ns

   ›Reveal solutionSolution

   Impulse is the time-integral of the force; integrating the given force profile from $t=0$ to the moment the force vanishes gives $0.9$ Ns — option (C).

   Concept and Intuition

   Impulse equals the area under a force-vs-time graph, i.e. $J=\int F\,dt$. Here the force decreases linearly with time and reaches zero exactly when the bullet leaves the barrel, so that instant is the natural upper limit of integration.

   Step-by-Step Solution

   1. Find when the force becomes zero: $600-2\times10^5t_0=0 \Rightarrow t_0=\dfrac{600}{2\times10^5}=3\times10^{-3}\ \mathrm s$.
   2. Impulse imparted while the bullet is in the barrel: $J=\displaystyle\int_0^{t_0}(600-2\times10^5t)\,dt$.
   3. Integrate: $J=\Big[600t-10^5t^2\Big]_0^{t_0}$.
   4. Evaluate: $600(3\times10^{-3})=1.8$; and $10^5(3\times10^{-3})^2=10^5\times9\times10^{-6}=0.9$.
   5. So $J=1.8-0.9=0.9\ \mathrm{Ns}$.

   Common Mistakes

   • Using the wrong upper limit (e.g. integrating over an arbitrary or infinite time) instead of the instant the force genuinely becomes zero.
   • Forgetting the factor of $10^5$ when integrating $-2\times10^5t$, i.e. writing $-2\times10^5t^2$ instead of $-10^5t^2$ (missing the $\tfrac12$ from integration).

   ✓Final answer

   The correct option is (C) — 0.9 Ns.

   ANSWER: C

5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.In a cricket match, bowler throws a ball of mass 0.2 kg with a speed of 72 kmph. The batsman deflected the ball by an angle of $45^0$ without changing its initial speed. The impulse imparted to the ball is? ($\mathrm{Cos}\ 22.5^0 = 0.92$) (A) 7.4 kg $\mathrm{ms^{-1}}$ (B) 8.2 kg $\mathrm{ms^{-1}}$ (C) 9.2 kg $\mathrm{ms^{-1}}$ (D) 8.4 kg $\mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   Using the impulse formula for a ball deflected by angle $\theta$ from its reverse path at unchanged speed, $|\Delta p| = 2mv\cos(\theta/2)$, gives 7.4 kg·m/s.

   Concept and Intuition

   When a ball bounces off a bat and its speed is unchanged but its direction is turned through some angle relative to a direct "bounce straight back" path, the impulse (change in momentum) can be found geometrically: the initial and final momentum vectors have equal magnitude $mv$, and the angle between the final velocity and the reverse of the initial velocity is the given deflection angle $\theta = 45^\circ$. Using the isosceles-triangle (vector diagram) relation for two equal-magnitude vectors separated by angle $(180^\circ-\theta)$ from each other:

   $|\Delta p|^2 = 2(mv)^2\big(1-\cos(180^\circ-\theta)\big) = 2(mv)^2(1+\cos\theta) = 4(mv)^2\cos^2(\theta/2)$

   $\implies |\Delta p| = 2mv\cos(\theta/2)$

   Step-by-Step Solution

   1. Convert speed: $v = 72\ \text{kmph} = 72\times\frac{5}{18} = 20\ \mathrm{ms^{-1}}$.
   2. Mass $m=0.2$ kg, deflection angle $\theta=45^\circ$, so $\theta/2 = 22.5^\circ$, with $\cos22.5^\circ = 0.92$ (given).
   3. Apply the impulse formula:

   $|\Delta p| = 2mv\cos(\theta/2) = 2(0.2)(20)(0.92) = 8 \times 0.92 = 7.36\ \text{kg}\,\mathrm{ms^{-1}} \approx 7.4\ \text{kg}\,\mathrm{ms^{-1}}$

   Common Mistakes

   • Using $\sin(\theta/2)$ instead of $\cos(\theta/2)$ — this happens if the deflection angle is (mis)interpreted as the angle between the initial and final directions directly, rather than the deviation from a straight bounce-back. The given hint value $\cos22.5^\circ$ signals which form to use.
   • Forgetting to convert kmph to m/s before substituting.

   ✓Final answer

   The correct option is (A) — 7.4 kg $\mathrm{ms^{-1}}$.

   ANSWER: A

6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The linear momentum of a body of mass 8 kg is $24\ \mathrm{kg\,m\,s^{-1}}$. If a constant force of 24 N acts on the body in the direction of motion of the body for a time of 3 s, then the increase in the kinetic energy of the body is (A) 480 J (B) 540 J (C) 270 J (D) 240 J

   ›Reveal solutionSolution

   This tests using impulse to find final momentum, then computing the change in kinetic energy; the answer is (B).

   Concept and Intuition

   When a constant force acts along the direction of motion, the impulse it delivers ($F\times t$) adds directly to the existing momentum. Once you know initial and final momenta, kinetic energies follow from $KE=\dfrac{p^2}{2m}$ or $\dfrac12mv^2$.

   Step-by-Step Solution

   1. Initial momentum $p_i=24\ \mathrm{kg\,m/s}$, mass $m=8$ kg, so initial velocity $v_i=\dfrac{p_i}{m}=\dfrac{24}{8}=3\ \mathrm{m/s}$.
   2. Initial KE: $KE_i=\dfrac12mv_i^2=\dfrac12(8)(3)^2=4\times9=36$ J.
   3. Impulse delivered by the 24 N force over 3 s: $J=Ft=24\times3=72\ \mathrm{kg\,m/s}$.
   4. Final momentum: $p_f=p_i+J=24+72=96\ \mathrm{kg\,m/s}$, so final velocity $v_f=\dfrac{96}{8}=12\ \mathrm{m/s}$.
   5. Final KE: $KE_f=\dfrac12(8)(12)^2=4\times144=576$ J.
   6. Increase in KE $=KE_f-KE_i=576-36=540$ J.

   Common Mistakes

   • Using the impulse-momentum theorem incorrectly (e.g., forgetting the force acts along the direction of existing motion, so momenta simply add).
   • Computing work done by the force via $Fd$ using average velocity but making an arithmetic slip — the momentum approach above avoids this.

   ✓Final answer

   The correct option is (B) — 540 J.

   ANSWER: B

7. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Two balls each of mass 250 g moving in opposite directions each with a speed $16\ \mathrm{m\,s^{-1}}$ collide and rebound with the same speeds. The impulse imparted to one ball due to the other is (A) $4\ \mathrm{kg\,m\,s^{-1}}$ (B) $16\ \mathrm{kg\,m\,s^{-1}}$ (C) $8\ \mathrm{kg\,m\,s^{-1}}$ (D) $2\ \mathrm{kg\,m\,s^{-1}}$

   ›Reveal solutionSolution

   Impulse on a ball equals its change in momentum; since velocity reverses direction, $\Delta v=32$ m/s, giving impulse $8\ \mathrm{kg\,m\,s^{-1}}$.

   Concept and Intuition

   Impulse-momentum theorem: the impulse delivered to a ball equals its change in momentum, $J=m(v_f-v_i)$. Rebounding at the same speed but reversed direction means the velocity doesn't just stop — it flips sign, so the change is twice the speed, not just the speed itself.

   Step-by-Step Solution

   1. Take one ball's initial velocity as $+16\ \mathrm{m/s}$ (before collision).
   2. After the elastic-like rebound, its velocity becomes $-16\ \mathrm{m/s}$ (same speed, opposite direction).
   3. Change in velocity: $\Delta v = (-16)-(+16)=-32\ \mathrm{m/s}$, magnitude $32\ \mathrm{m/s}$.
   4. Mass $m=250\ \mathrm{g}=0.25\ \mathrm{kg}$.
   5. Impulse magnitude: $J=m|\Delta v|=0.25\times32=8\ \mathrm{kg\,m\,s^{-1}}$.

   Common Mistakes

   • Using $\Delta v=16$ m/s (just the speed) instead of $32$ m/s, forgetting the direction reversal doubles the change.
   • Confusing impulse on one ball with the (zero) net change of the two-ball system's total momentum.

   ✓Final answer

   The correct option is (C) — $8\ \mathrm{kg\,m\,s^{-1}}$.

   ANSWER: C

8. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Water flowing through a pipe of area of cross-section $2\times10^{-3}\ \mathrm{m^2}$ hits a vertical wall horizontally with a velocity of $12\ \mathrm{m\,s^{-1}}$. If the water does not rebound after hitting the wall, then the force acting on the wall due to water is (A) 24 N (B) 144 N (C) 288 N (D) 72 N

   ›Reveal solutionSolution

   Tests momentum flux of a fluid jet hitting a wall with no rebound. Answer: 288 N.

   Concept and Intuition

   When a steady stream of fluid strikes a surface and does not bounce back, the fluid's momentum along the direction of flow is annihilated at the wall. By Newton's second law applied to the fluid (rate of change of momentum = force), the wall must supply a force equal in magnitude to the rate at which momentum arrives — and by Newton's third law, the water pushes back on the wall with the same magnitude.

   Step-by-Step Solution

   1. Mass flow rate of water: $\dot{m} = \rho A v$, where $\rho = 1000\ \mathrm{kg\,m^{-3}}$ is the density of water, $A = 2\times10^{-3}\ \mathrm{m^2}$ is the cross-sectional area, and $v = 12\ \mathrm{m\,s^{-1}}$ is the speed.

   $\dot{m} = 1000 \times 2\times10^{-3} \times 12 = 24\ \mathrm{kg\,s^{-1}}$

   2. Since the water does not rebound, its horizontal velocity drops from $v$ to $0$ at the wall — the change in velocity per unit mass is $v$ itself (not $2v$, which would apply for elastic rebound).
   3. Rate of change of momentum (= force on water, and equal and opposite force on wall):

   $F = \dot{m}\, v = 24 \times 12 = 288\ \mathrm{N}$

   Common Mistakes

   • Using $2v$ instead of $v$ — that formula is for perfectly elastic rebound, not for water that sticks/splatters without bouncing back.
   • Forgetting to convert area units or misplacing the density value.

   ✓Final answer

   The correct option is (C) — 288 N.

   ANSWER: C

9. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.A disc of mass 0.2 kg is kept floating in air without falling by vertically firing bullets each of mass 0.05 kg on the disc at the rate of 10 bullets per every second. If the bullets rebound with the same speed, then the speed of each bullet is (Acceleration due to gravity $= 10 \mathrm{~ms^{-2}}$) (A) $2 \mathrm{~m~s^{-1}}$ (B) $10 \mathrm{~m~s^{-1}}$ (C) $20 \mathrm{~m~s^{-1}}$ (D) $1 \mathrm{~m~s^{-1}}$

   ›Reveal solutionSolution

   The average force from the rebounding bullets must balance the disc's weight; each rebound transfers momentum $2m_bv$, and with 10 bullets/s this gives $v=2$ m/s.

   Concept and Intuition

   Even though bullets hit and bounce off individually (a series of discrete impacts), if enough of them strike per second, the time-averaged force they exert behaves like a continuous force (this is the same idea used for rocket thrust and for pressure from gas molecules). For a rebounding ("elastic", same-speed reversal) collision, each bullet's momentum changes by $2m_bv$ (from $+m_bv$ to $-m_bv$), and by Newton's third law the disc receives that momentum kick upward.

   Step-by-Step Solution

   1. For the disc to float (stay in equilibrium), the average upward force from the bullets must equal its weight:

   $F=Mg=0.2\times10=2\ \text{N}.$

   2. Each bullet's momentum changes from $+m_bv$ (incoming) to $-m_bv$ (rebounding at the same speed), a change of magnitude $\Delta p=2m_bv$.
   3. With $n=10$ bullets striking per second, the average force delivered to the disc is

   $F=n\times\Delta p = n\times2m_bv = 10\times2\times0.05\times v = v.$

   4. Equating to the required force: $v=2$ m/s.

   Common Mistakes

   • Using $\Delta p=m_bv$ (as if the bullet simply stops) instead of $2m_bv$ (since it rebounds with the same speed, reversing direction) — this would give a wrong factor-of-2 answer.
   • Forgetting to multiply by the rate $n=10$ bullets/s to get force from momentum-change-per-bullet.

   ✓Final answer

   The correct option is (A) — $2\ \mathrm{m~s^{-1}}$.

   ANSWER: A

10. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If a body of mass 2 kg moving with initial velocity of $4\ \mathrm{m\,s^{-1}}$ is subjected to a force of 3 N for a time of 2 s normal to the direction of its initial velocity, then the resultant velocity of the body is (A) $7\ \mathrm{m\,s^{-1}}$ (B) $5\ \mathrm{m\,s^{-1}}$ (C) $2\ \mathrm{m\,s^{-1}}$ (D) $7.5\ \mathrm{m\,s^{-1}}$

    ›Reveal solutionSolution

    A force applied perpendicular to a moving body's velocity adds a new independent velocity component; the two combine as a right-angle vector sum, giving 5 m/s.

    Concept and Intuition

    Force (and hence acceleration and impulse) components act independently along their own directions. A force perpendicular to the existing velocity does not change the original component's magnitude — it only adds a new component in its own direction. The two components then combine vectorially (Pythagoras) for the resultant speed.

    Step-by-Step Solution

    1. Original velocity: $v_1 = 4\ \mathrm{m\,s^{-1}}$ (unchanged, since the force has no component along this direction).
    2. Impulse in the perpendicular direction: $J = Ft = 3\times2 = 6\ \mathrm{kg\,m\,s^{-1}}$.
    3. New perpendicular velocity component: $v_2 = J/m = 6/2 = 3\ \mathrm{m\,s^{-1}}$.
    4. Resultant velocity: $v = \sqrt{v_1^2+v_2^2} = \sqrt{16+9} = \sqrt{25} = 5\ \mathrm{m\,s^{-1}}$.

    Common Mistakes

    • Adding the velocities directly (4+3=7) instead of vectorially, since they are perpendicular, not along the same line.
    • Forgetting the original velocity component is unaffected by a purely perpendicular force.

    ✓Final answer

    The correct option is (B) — $5\ \mathrm{m\,s^{-1}}$.

    ANSWER: B

11. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.A man of mass 70 kg jumps to a height of 0.8 m from the ground, then the momentum transferred by the ground to man is $(g = 10\ \text{ms}^{-2})$ (A) $280\ \text{kg ms}^{-1}$ (B) $200\ \text{kg ms}^{-1}$ (C) $560\ \text{kg ms}^{-1}$ (D) $400\ \text{kg ms}^{-1}$

    ›Reveal solutionSolution

    Find the take-off speed needed to reach height 0.8 m using $v=\sqrt{2gh}$, then the momentum imparted by the ground equals $mv$.

    Concept and Intuition

    To jump to a certain height, a person must leave the ground with just enough vertical speed that, under gravity alone, they rise to that height. By Newton's third law, the ground pushes back on the person with exactly the impulse needed to change their momentum from zero (standing) to $mv$ (at the instant of leaving the ground).

    Step-by-Step Solution

    1. Find the required take-off speed from $v^2 = 2gh$: $v = \sqrt{2\times10\times0.8} = \sqrt{16} = 4\ \text{m/s}$.
    2. The man's momentum changes from $0$ (before push-off) to $mv$ (at lift-off).
    3. Impulse (momentum transferred) $= mv - 0 = 70\times4 = 280\ \text{kg m/s}$.

    Common Mistakes

    • Trying to include the round trip (jump up and fall back) — the question only asks about momentum transferred during push-off, i.e. up to the point of leaving the ground.
    • Arithmetic slip in $\sqrt{2gh}$: forgetting to take the square root, or using wrong $g$.

    ✓Final answer

    The correct option is (A) — 280 kg ms⁻¹.

    ANSWER: A

12. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.A block of metal 4 kg is in rest on a frictionless surface. It was targeted by a jet releasing water of $2\ \mathrm{kg\,s^{-1}}$ at a speed of $10\ \mathrm{ms^{-1}}$. The acceleration of the block is (A) $10\ \mathrm{ms^{-2}}$ (B) $15\ \mathrm{ms^{-2}}$ (C) $20\ \mathrm{ms^{-2}}$ (D) $5\ \mathrm{ms^{-2}}$

    ›Reveal solutionSolution

    A water jet hitting a block and splattering (not rebounding) exerts $F=\dot m v$; dividing by the block's mass gives its acceleration, $5\ \mathrm{ms^{-2}}$.

    Concept and Intuition

    When a continuous stream of water strikes a surface and simply spreads out sideways (as opposed to bouncing straight back), each parcel of water loses all its forward momentum on impact. By Newton's second law in the form $F=\dfrac{dp}{dt}$, the rate at which momentum is delivered to the block equals the force on it: $F=\dot m\,v$, where $\dot m$ is the mass flow rate and $v$ its incoming speed.

    Step-by-Step Solution

    1. Mass flow rate $\dot m = 2\ \mathrm{kg\,s^{-1}}$, jet speed $v=10\ \mathrm{ms^{-1}}$.
    2. Water does not bounce back (it "releases" onto/around the block), so momentum change per unit time $=\dot m v - 0 = \dot m v$.
    3. Force on block: $F=\dot m v = 2\times10=20\ \mathrm N$.
    4. Block mass $m=4$ kg (frictionless surface, so this is the net force).
    5. Acceleration: $a=F/m=20/4=5\ \mathrm{ms^{-2}}$.

    Common Mistakes

    • Doubling the force assuming an elastic bounce-back (that would give $10\ \mathrm{ms^{-2}}$, not asked here since the water "releases" onto the block, not rebounds).
    • Forgetting to divide by the block's mass and reporting the force as the acceleration.

    ✓Final answer

    The correct option is (D) — $5\ \mathrm{ms^{-2}}$.

    ANSWER: D