Q.A shell of mass $0.020\ \text{kg}$ is fired by a gun of mass $100\ \text{kg}$. If the muzzle speed of the shell is $80\ \text{m s}^{-1}$, what is the recoil speed of the gun?

Showing the 12 most recent of 31 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Three balls A, B and C of equal mass are moving in the same order along the same straight line on a smooth horizontal surface. Initially, ball A moves towards right with a velocity of $6\ ms^{-1}$, ball B moves towards left with a velocity of $4\ ms^{-1}$ and ball C moves towards right with a velocity of $3\ ms^{-1}$. If the coefficient of restitution between any two balls is 0.8, then the relative velocity of balls B and C after collision is (A) $1.6\ ms^{-1}$ (B) $0.8\ ms^{-1}$ (C) $3.2\ ms^{-1}$ (D) $2.4\ ms^{-1}$

   ›Reveal solutionSolution

   A-B collide first, then the faster rebounding B catches C; the restitution relation directly gives their post-collision relative velocity of 1.6 m/s.

   Concept and Intuition

   For a 1-D collision between equal masses with restitution $e$, the relative velocity of separation always equals $e$ times the relative velocity of approach — this is the definition of $e$, and it applies to whichever pair collides, in sequence.

   Step-by-Step Solution

   1. Take rightward as positive: $v_A=6$, $v_B=-4$, $v_C=3$. A (moving right) and B (moving left) approach each other first; C moves away from B, so it isn't involved yet.
   2. A-B collision (equal masses, $e=0.8$): momentum conservation $v_A'+v_B'=v_A+v_B=2$; restitution $v_B'-v_A'=e(v_A-v_B)=0.8(6-(-4))=8$.
   3. Solving: $v_B'=\dfrac{2+8}{2}=5$, $v_A'=\dfrac{2-8}{2}=-3$.
   4. Now $v_A'=-3$ (moving away, no further collision), $v_B'=5>v_C=3$: B is behind C and now catches up, so B and C collide next. …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A body which is initially at rest breaks into 2 pieces of masses 4M and 6M respectively, together having a total kinetic energy E. The piece with mass 4M, after breaking has a kinetic energy (A) 0.6 E (B) 0.4 E (C) 0.2 E (D) 0.8 E

   ›Reveal solutionSolution

   This tests momentum conservation in an explosion from rest, combined with the fact that kinetic energy for a given momentum scales inversely with mass. The answer is (A), 0.6E.

   Concept and Intuition

   When a body at rest breaks into two pieces, total momentum must remain zero, so the two pieces fly apart with equal and opposite momenta ($p$ and $-p$). Since $KE=\dfrac{p^2}{2m}$, for a fixed momentum magnitude the lighter piece gets more kinetic energy than the heavier piece.

   Step-by-Step Solution

   1. Momentum conservation (initial momentum zero): $4M v_1 = 6M v_2$ in magnitude, i.e. the momenta are equal and opposite: $p_1=|4Mv_1|=p_2=|6Mv_2|=p$.
   2. Kinetic energies: $KE_{4M}=\dfrac{p^2}{2(4M)}$, $KE_{6M}=\dfrac{p^2}{2(6M)}$.
   3. Ratio: $\dfrac{KE_{4M}}{KE_{6M}}=\dfrac{6M}{4M}=\dfrac{3}{2}$.
   4. Let $KE_{6M}=x$; then $KE_{4M}=1.5x$. Total: $1.5x+x=2.5x=E \Rightarrow x=0.4E$. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A moving block having mass m, collides with another stationary block of mass 5m. After collision, the block with mass m comes to rest. If the initial velocity of block with mass m is V, then the value of coefficient of restitution is (A) 0.2 (B) 0.5 (C) 0.7 (D) 0.25

   ›Reveal solutionSolution

   This tests applying momentum conservation together with the definition of the coefficient of restitution for a one-dimensional collision. The answer is (A), 0.2.

   Concept and Intuition

   The coefficient of restitution $e$ is the ratio of the relative velocity of separation to the relative velocity of approach. To find it, we first need the final velocities of both bodies, which requires momentum conservation since we're told one final velocity (block $m$ stops) but not the other.

   Step-by-Step Solution

   1. Before collision: block $m$ moves with velocity $V$, block $5m$ is at rest.
   2. Momentum conservation: $mV+5m(0)=m(0)+5m\,v_2'$, so $mV=5m\,v_2' \Rightarrow v_2'=\dfrac{V}{5}$.
   3. Relative velocity of approach $=V-0=V$.
   4. Relative velocity of separation $=v_2'-v_1'=\dfrac{V}{5}-0=\dfrac{V}{5}$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A Ball 'A' collides with another identical ball 'B' which is at rest. After collision, if the velocity of the ball 'B' becomes two times of final velocity of the ball 'A', then the coefficient of restitution is (A) $\dfrac{1}{3}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   Using momentum conservation to relate the final speeds ($v_B=2v_A$) and then applying the restitution formula $e=\frac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$ gives $e=\tfrac13$.

   Concept and Intuition

   A collision problem with unknown final speeds needs two equations: conservation of momentum (always true) and the definition of the coefficient of restitution (which encodes the given condition here). Since we're told a ratio between the final speeds, momentum conservation alone can pin down both speeds in terms of the initial speed.

   Step-by-Step Solution

   1. Let the identical balls have mass $m$. Ball A moves with initial speed $u$; ball B is initially at rest.
   2. Let final speeds be $v_A$ (for A) and $v_B$ (for B), with the given condition $v_B=2v_A$.
   3. Conservation of momentum: $mu+0=mv_A+mv_B \Rightarrow u=v_A+v_B=v_A+2v_A=3v_A$.
   4. So $v_A=\dfrac{u}{3}$ and $v_B=2v_A=\dfrac{2u}{3}$. …

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5. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A man weighing 80 kg is standing on a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a speed of $1\ \text{ms}^{-1}$, then after 4 seconds his displacement relative to the ground will be (A) 5 m (B) 4.8 m (C) 3.2 m (D) 3.0 m

   ›Reveal solutionSolution

   Conservation of momentum on the man+trolley system (net momentum stays zero)

   gives the man's ground speed as 0.8 m/s, so his displacement in 4 s is

   3.2 m. Answer: (C).

   Concept and Intuition

   The man and trolley form an isolated system on frictionless rails — the net

   external horizontal force is zero, so total momentum is conserved. Since the

   system starts at rest, the total momentum stays exactly zero at all times:

   whenever the man moves one way, the trolley must move the opposite way to

   keep the total momentum zero.

   Step-by-Step Solution

   1. Let $v_m$ = man's velocity (ground frame), $v_t$ = trolley's velocity (ground frame), both taken along the direction the man walks.
   2. Momentum conservation (starting from rest): $m\,v_m + M\,v_t = 0$, i.e. $80\,v_m + 320\,v_t = 0 \Rightarrow v_m = -4v_t$.
   3. Relative velocity of man w.r.t. trolley is given as $1\ \text{ms}^{-1}$: $v_m - v_t = 1$. …

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6. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A & B are two similar balls. Ball A hits directly ball B, which is at rest. After impact, the ball A comes to rest. If half of the kinetic energy is lost in the collision, the coefficient of restitution is (A) 0.8 (B) 1 (C) 0.5 (D) 0.7

   ›Reveal solutionSolution

   All the surviving kinetic energy is carried by ball $B$; with half the KE lost, the separation speed is $u/\sqrt2$, giving $e=\dfrac{1}{\sqrt2}\approx 0.7$.

   Ball $A$ strikes stationary ball $B$ head-on with speed $u$ and comes to rest after impact, so all the final kinetic energy resides in ball $B$ (speed $v_B$).

   Energy condition. Half the kinetic energy is lost, so the final KE is half the initial KE:

   $\frac{1}{2}m v_B^{2}=\frac{1}{2}\left(\frac{1}{2}m u^{2}\right)\;\Rightarrow\;v_B^{2}=\frac{u^{2}}{2}\;\Rightarrow\;v_B=\frac{u}{\sqrt2}.$ …

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7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Two identical balls $P$ and $Q$ are having velocities $0.7\,\mathrm{ms^{-1}}$ and $-0.4\,\mathrm{ms^{-1}}$ respectively are colliding in one dimension elastically. The velocities of $P$ and $Q$ after the collision respectively are (A) $0.7\,\mathrm{ms^{-1}}, -0.4\,\mathrm{ms^{-1}}$ (B) $-0.4\,\mathrm{ms^{-1}}, 0.7\,\mathrm{ms^{-1}}$ (C) $+0.4\,\mathrm{ms^{-1}}, -0.7\,\mathrm{ms^{-1}}$ (D) $+0.7\,\mathrm{ms^{-1}}, -0.7\,\mathrm{ms^{-1}}$

   ›Reveal solutionSolution

   Equal-mass elastic collision in 1D: velocities are exchanged, so $P\to-0.4\,\mathrm{ms^{-1}}$ and $Q\to0.7\,\mathrm{ms^{-1}}$.

   Concept and Intuition

   For a 1D elastic collision between two particles of equal mass $m$, both momentum and kinetic energy are conserved. Solving the two conservation equations for equal masses gives the elegant result that the particles simply swap velocities — this is a standard, memorable consequence worth deriving once.

   Step-by-Step Solution

   1. Let $u_P=0.7\,\mathrm{ms^{-1}}$, $u_Q=-0.4\,\mathrm{ms^{-1}}$, masses equal ($m$).
   2. Momentum conservation: $mu_P+mu_Q=mv_P+mv_Q \Rightarrow u_P+u_Q=v_P+v_Q$.
   3. For elastic collisions, relative velocity of separation = relative velocity of approach (reversed): $v_Q-v_P = -(u_Q-u_P) = u_P-u_Q$.
   4. Adding/subtracting these two equations for equal masses gives $v_P=u_Q$ and $v_Q=u_P$. …

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8. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A block of mass 10 kg moving with a speed of $5\hat{i}\ \mathrm{m\,s^{-1}}$ on a frictionless horizontal surface suddenly explodes into two pieces. If one piece with mass 4 kg moves with a speed of $10\hat{i}\ \mathrm{m\,s^{-1}}$, then the velocity of the second piece is (A) $7.67\ \mathrm{m\,s^{-1}}$ (B) $1.67\ \mathrm{m\,s^{-1}}$ (C) 6.67s m s$^{-1}$ (D) $2.67\ \mathrm{m\,s^{-1}}$

   ›Reveal solutionSolution

   Momentum conservation during the explosion gives the second piece a velocity of $1.67\ \mathrm{m\,s^{-1}}$ in the same direction, option (B).

   Concept and Intuition

   An internal explosion involves only internal forces between the two fragments, so the total momentum of the system just before and just after the explosion must be equal (Newton's third law guarantees the internal impulses cancel). We only need the masses and one fragment's velocity to find the other.

   Step-by-Step Solution

   1. Total mass before explosion: 10 kg, moving at $5\hat i\ \mathrm{m\,s^{-1}}$.
   2. Initial momentum: $p_i=10\times5=50$ kg·m/s (along $\hat i$).
   3. After explosion: piece 1 has mass 4 kg moving at $10\hat i\ \mathrm{m\,s^{-1}}$, so its momentum is $p_1=4\times10=40$ kg·m/s.
   4. Piece 2 has mass $10-4=6$ kg. By conservation of momentum, $p_1+p_2=p_i$, so $p_2=50-40=10$ kg·m/s.
   5. Velocity of piece 2: $v_2=p_2/m_2=10/6=1.67\ \mathrm{m\,s^{-1}}$ (along $\hat i$, same direction as the original motion). …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A steel sphere of radius 1.2 cm collides a second steel sphere at rest. If the collision is elastic and after the collision the first sphere continues to move in its initial direction with a velocity of $\dfrac{7}{9}$ times its initial velocity, then the radius of the second sphere is (A) 1.8 cm (B) 2.4 cm (C) 1.2 cm (D) 0.6 cm

   ›Reveal solutionSolution

   Using the elastic-collision velocity formula and mass $\propto r^3$ for spheres of the same material, the second sphere's radius comes out to (D) 0.6 cm.

   Concept and Intuition

   In a 1-D elastic collision where the target is initially at rest, the incident body's velocity after collision is $v_1'=\dfrac{m_1-m_2}{m_1+m_2}v_1$ — this ratio depends only on the mass ratio. Since both spheres are made of steel (same density), their masses scale with volume, i.e. with $r^3$. So once we find the mass ratio from the velocity condition, we can directly get the radius ratio by taking a cube root.

   Step-by-Step Solution

   1. Elastic collision formula (target initially at rest): $v_1'=\dfrac{m_1-m_2}{m_1+m_2}v_1$.
   2. Given $v_1'=\dfrac79v_1$ (same direction, so the ratio is positive, meaning $m_1>m_2$):

   $\frac{m_1-m_2}{m_1+m_2}=\frac79.$

   3. Cross-multiply: $9(m_1-m_2)=7(m_1+m_2)\ \Rightarrow\ 9m_1-9m_2=7m_1+7m_2\ \Rightarrow\ 2m_1=16m_2\ \Rightarrow\ m_1=8m_2$. …

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10. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
    Q.A body A of mass 200 g moving with a velocity $v_1\hat{i}$ makes collision with another body B of mass 100 g moving with a velocity $v_2\hat{i}$. After collision, A and B move with velocities $v_3\hat{i}$ and $v_4\hat{i}$ respectively. If $v_3 = 0.5v_1$, then the value of $v_1$ is (A) $v_4 - \dfrac{v_2}{4}$ (B) $v_4 - \dfrac{v_2}{2}$ (C) $v_4 - v_2$ (D) $v_4 + v_2$

    ›Reveal solutionSolution

    This tests conservation of linear momentum in a one-dimensional collision. The answer is (C) $v_4 - v_2$.

    Concept and Intuition

    In any collision (elastic or inelastic) with no external horizontal forces, the total linear momentum of the system before and after collision must be equal. Here, expressing this conservation law algebraically and substituting the given relationship between the post-collision velocities lets us solve directly for $v_1$ in terms of the other velocities.

    Step-by-Step Solution

    1. Masses: $m_A = 200\text{ g} = 0.2\text{ kg}$, $m_B = 100\text{ g} = 0.1\text{ kg}$.
    2. Momentum conservation: $m_A v_1 + m_B v_2 = m_A v_3 + m_B v_4$, i.e. $0.2v_1 + 0.1v_2 = 0.2v_3 + 0.1v_4$.
    3. Given $v_3 = 0.5v_1$, substitute: $0.2v_1 + 0.1v_2 = 0.2(0.5v_1) + 0.1v_4 = 0.1v_1 + 0.1v_4$. …

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11. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.A bomb of mass 12 kg at rest explodes into two pieces of masses 4 kg and 8 kg. If the velocity of 8 kg mass is $6 \mathrm{~m~s^{-1}}$, then the kinetic energy of 4 kg mass is (A) 248 J (B) 326 J (C) 124 J (D) 288 J

    ›Reveal solutionSolution

    Momentum conservation from an initially-at-rest bomb gives the 4 kg fragment's speed as 12 m/s, so its kinetic energy is 288 J.

    Concept and Intuition

    A bomb at rest has zero total momentum. When it explodes into two pieces, the internal explosive forces are equal and opposite (Newton's third law), so the two fragments must fly off with equal and opposite momenta to keep total momentum zero (momentum conservation, no external force during the near-instantaneous explosion).

    Step-by-Step Solution

    1. Initial momentum = 0 (bomb at rest).
    2. After explosion: $m_1 v_1 = m_2 v_2$ in magnitude (opposite directions), with $m_1=4\,\mathrm{kg}$, $m_2=8\,\mathrm{kg}$, $v_2=6\,\mathrm{m/s}$.
    3. Solve for $v_1$: $v_1 = \dfrac{m_2 v_2}{m_1} = \dfrac{8\times 6}{4} = 12\ \mathrm{m\,s^{-1}}$. …

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12. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.A body moving along a straight line collides another body of same mass moving in the same direction with half of the velocity of the first body. If the coefficient of restitution between the two bodies is 0.5, then the ratio of the velocities of the two bodies after collision is (treat the collision as one dimensional) (A) $2:5$ (B) $2:3$ (C) $5:7$ (D) $3:7$

    ›Reveal solutionSolution

    This tests one-dimensional collisions using both momentum conservation and the coefficient of restitution; the answer is (C) $5:7$.

    Concept and Intuition

    For a collision that is neither perfectly elastic nor perfectly inelastic, two equations govern the outcome: conservation of linear momentum, and Newton's experimental law of restitution, $e = \dfrac{\text{relative velocity of separation}}{\text{relative velocity of approach}}$. Together with equal masses, these two linear equations in $v_1, v_2$ can be solved directly.

    Step-by-Step Solution

    1. Let the first body have initial velocity $u_1 = u$ and the second (moving in the same direction, ahead of it) $u_2 = u/2$; both have mass $m$.
    2. Momentum conservation: $mu_1 + mu_2 = mv_1 + mv_2 \Rightarrow v_1+v_2 = u + \tfrac{u}{2} = \tfrac{3u}{2}$.
    3. Restitution equation: $v_2 - v_1 = e(u_1-u_2) = 0.5\left(u - \tfrac{u}{2}\right) = \tfrac{u}{4}$. …

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