Q.A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m/s. If the mass of ball is .15 kg, determine the impulse imparted to the ball (Assume Linear Motion of the ball). OR State the Law of Conservation of Linear Momentum. Explain it by giving any one example.

Showing the 12 most recent of 24 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.What is that physical quantity called which is equal to the change in momentum of a body?

   (a) Force

   (b) Acceleration

   (c) Impulse

   (d) Reaction

   ›Reveal solutionSolution

   By definition, impulse J = change in momentum = delta p = F x delta t.

   Starting from Newton's second law in its more general (momentum) form:

   F = dp/dt

   Integrating both sides over the time interval the force acts:

   J = Integral(F dt) = delta p = p_final - p_initial

   …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Dimensional formula of impulse is similar to the dimensional formula of -

   (a) Force

   (b) Angular momentum

   (c) Pressure

   (d) Linear momentum

   ›Reveal solutionSolution

   Impulse J = FΔt has the same dimensional formula as linear momentum, MLT⁻¹, because the impulse-momentum theorem states J = Δp.

   Impulse is defined as J = F × Δt, dimensions [MLT⁻²][T] = MLT⁻¹. Linear momentum is p = mv, dimensions [M][LT⁻¹] = MLT⁻¹ — identical. This is not a coincidence: Newton's second law states F = dp/dt, so integrating over time, F·Δt = Δp, i.e., impulse always equals the change in linear momentum (the impulse-mo …

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3. CBSE 2026Set sz1 markMCQ
   Q.An impulse is the change in:

   (a) velocity

   (b) kinetic energy

   (c) momentum

   (d) displacement

   ›Reveal solutionSolution

   Impulse J = F(avg) x delta t = change in momentum (delta p) of the body.

   From Newton's second law, F = dp/dt, so integrating force over the short time of contact: J = integral F dt = delta p = mv(final) - mv(initial). Impulse and change in momentum h …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the column - select the correct definition (from Column B) for the term 'Impulse' (Column A):

   (a) change in linear momentum

   (b) motion opposing force

   (c) loss of energy

   (d) rate of change of momentum

   (e) ability of doing work

   (f) rate of doing work

   ›Reveal solutionSolution

   Impulse equals the change produced in linear momentum.

   Impulse J is defined as J = F × Δt for a constant force acting over a short time Δt (or ∫F dt for a varying force). Since F = dp/dt, integrating over the interval of action gives:

   J = ∫ F dt = ∫ dp = Δp = p_final − p_initial …

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5. CBSE 2026Set ANN1 markMCQ
   Q.A graph is drawn with time along x axis and force along y axis. The area under the graph represents

   (a) torque

   (b) impulse

   (c) momentum

   (d) couple

   ›Reveal solutionSolution

   The area under a force-time graph is impulse, which also equals the change in momentum.

   Impulse is defined as the product of force and the time for which it acts: J = integral of F dt. On a graph with time on the x-axis and force on the y-axis, this integral is precisely the area enclosed between the curve and the time axis.

   By the impulse-momentum theorem, this impulse also equals the change in the body's linear momentum: …

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6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in Momentum of the particle over the time interval from zero to 8 second is:

   (a) 24 Ns

   (b) 20 Ns

   (c) 12 Ns

   (d) 6 Ns

   ›Reveal solutionSolution

   Impulse = change in momentum = area under the F–t graph (areas below the axis count as negative). Summing the three regions of this graph gives 12 Ns.

   By the impulse–momentum theorem: $\Delta p = \int F\,dt =$ (net signed area under the F–t curve).

   Break the graph into its three described regions:

   Region 1 (0 to 2 s): F rises linearly from 0 to 6 N — a triangle.

   $\text{Area}_1 = (1/2)(2)(6) = 6$ Ns

   Region 2 (2 to 4 s): F is constant at $-3$ N — a rectangle below the axis.

   $\text{Area}_2 = (-3)(2) = -6$ Ns

   …

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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The momentum of a bus having mass 'm' standing at the bus stop is-

   (a) Infinite

   (b) Equal to its mass

   (c) Zero

   (d) None of the above

   ›Reveal solutionSolution

   The momentum of the standing bus is zero.

   Linear momentum of a body is defined as the product of its mass and velocity:

   $p = mv$

   …

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8. CBSE 2025Set ANNUAL1 mark
   Q.Name the physical quantity whose unit is Newton second.

   ›Reveal solutionSolution

   The quantity whose SI unit is the newton-second (N s) is impulse, defined as the product of force and the time for which it acts.

   Impulse $\vec J = \vec F\, \Delta t$ (or $\int \vec F\,dt$ for a varying force). Its unit is $N\cdot s = (\text{kg m s}^{-2})\cdot s = \text{kg m s}^{-1}$, the same as the unit of linear momentum — consistent with the impul …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.A body of mass m and speed v collides with a wall perpendicularly and returns with the same speed v. The momentum imparted to the wall is (A) mv (B) ½mv (C) 2mv (D) zero

   ›Reveal solutionSolution

   The wall receives momentum $2mv$ from the elastically bouncing body.

   Before collision the body's momentum (taking the incoming direction as positive) is $+mv$. After bouncing back with the same speed, its momentum is $-mv$. The change in the body's momentum is $\Delta p = (-mv) - (mv) = -2mv$, i.e. magnitude $2mv$.

   …

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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.A ball of mass m strikes a rigid wall with velocity u and rebounds with the same speed. The impulse imparted to the ball by the wall is

    (a) 2mu

    (b) mu

    (c) zero

    (d) -2mu

    ›Reveal solutionSolution

    Impulse = change in momentum. Since the ball's velocity reverses direction (same speed, opposite sign) after rebounding, the magnitude of the impulse works out to 2mu.

    Take the direction of the ball's motion toward the wall as positive.

    • Initial momentum, p_i = +mu
    • After rebounding with the same speed, the ball moves away from the wall: final momentum, p_f = -mu

    Impulse imparted to the ball = change in momentum = p_f - p_i = (-mu) - (mu) = -2mu

    …

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11. CBSE 2024Set ANNUAL1 markMCQ
    Q.Change in momentum is given by

    (a) Force x Mass

    (b) Force x Time

    (c) Force x Velocity

    (d) Force x Displacement

    ›Reveal solutionSolution

    Impulse (Force x Time) equals the change in momentum — this follows directly from Newton's second law.

    Newton's second law: F = dp/dt (rate of change of momentum). Rearranging for a constant force over a time interval Δt:

    Δp = F x Δt

    …

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12. CBSE 2024Set SET-NDP60001 mark
    Q.Change in momentum is always equal to ............ (force / impulse).

    ›Reveal solutionSolution

    The impulse–momentum theorem states that the change in momentum of a body equals the impulse applied to it: $\vec{J} = \Delta \vec{p}$.

    Impulse is defined as the product of force and the (short) time for which it acts: $\vec{J} = \vec{F}\,\Delta t$ for constant force, or $\vec{J} = \int \vec{F}\,dt$ in general. From Newton's second law, $\vec{F} = \dfrac{d\vec{p}}{dt}$, so

    $\vec{J} = \int \vec{F}\,dt = \int \dfrac{d\vec{p}}{dt}\,dt = \Delta \vec{p}$

    …

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