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Worked Examples · Example 4.4

Q.A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s−112\ \text{m s}^{-1}. If the mass of the ball is 0.15 kg0.15\ \text{kg}, determine the impulse imparted to the ball. (Assume linear motion of the ball.)

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The ball reverses direction at constant speed, so its velocity changes from +12 m s−1+12\ \text{m s}^{-1} to −12 m s−1-12\ \text{m s}^{-1}; impulse equals the change in momentum, giving 3.6 kg m s−1\boxed{3.6\ \text{kg m s}^{-1}} in magnitude.

Why impulse is about change in momentum

Impulse measures the effect of a force acting over time. Newton's second law in its most general form tells us that the net force equals the rate of change of momentum:

F⃗=dp⃗dt\vec{F} = \frac{d\vec{p}}{dt}

Integrating both sides over the collision time gives the impulse-momentum theorem:

J⃗=∫F⃗ dt=Δp⃗=p⃗final−p⃗initial\vec{J} = \int \vec{F}\, dt = \Delta \vec{p} = \vec{p}_{\text{final}} - \vec{p}_{\text{initial}}

The beauty here is that we don't need to know the force profile or contact time — only the momentum before and after.

Step-by-step calculation

  1. Set up a coordinate system.

    Choose the direction from bowler to batsman as positive. The ball initially travels toward the batsman at vi=+12 m s−1v_i = +12\ \text{m s}^{-1}.

  2. Find the initial momentum.

pi=mvi=0.15×12=1.8 kg m s−1p_i = m v_i = 0.15 \times 12 = 1.8\ \text{kg m s}^{-1}

  1. Determine the final velocity.

    The batsman hits the ball "straight back" at the same speed, so it now travels toward the bowler. In our coordinate system, vf=−12 m s−1v_f = -12\ \text{m s}^{-1}.

  2. Calculate the final momentum.

pf=mvf=0.15×(−12)=−1.8 kg m s−1p_f = m v_f = 0.15 \times (-12) = -1.8\ \text{kg m s}^{-1}

  1. Compute the impulse. J=pf−pi=−1.8−1.8=−3.6 kg m s−1J = p_f - p_i = -1.8 - 1.8 = -3.6\ \text{kg m s}^{-1} …

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