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NCERT Exemplar · Q3

Q.A cricket ball of mass 150 g has an initial velocity u=(3i^+4j^)\mathbf{u} = (3\hat{i} + 4\hat{j}) m s−1^{-1} and a final velocity v=−(3i^+4j^)\mathbf{v} = -(3\hat{i} + 4\hat{j}) m s−1^{-1} after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1^{1})

(a) zero
(b) −(0.45i^+0.6j^)-(0.45\hat{i} + 0.6\hat{j})
(c) −(0.9i^+1.2j^)-(0.9\hat{i} + 1.2\hat{j})
(d) −5(i^+j^)-5(\hat{i} + \hat{j}).
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The ball's velocity reverses direction completely, so the change in momentum is twice the initial momentum in the opposite direction: Δp=−(0.9i^+1.2j^)\Delta \mathbf{p} = -(0.9\hat{i} + 1.2\hat{j}) kg m s−1^{-1}.

When a cricket ball is struck, its momentum changes. Momentum is a vector quantity p=mv\mathbf{p} = m\mathbf{v}, and the change in momentum tells us about the impulse delivered by the bat. The key insight here is that the ball doesn't just stop—it reverses direction entirely, which means the momentum change is substantial.

The change in momentum is defined as:

Δp=pfinal−pinitial=mv−mu=m(v−u)\Delta \mathbf{p} = \mathbf{p}_{\text{final}} - \mathbf{p}_{\text{initial}} = m\mathbf{v} - m\mathbf{u} = m(\mathbf{v} - \mathbf{u})

This vector subtraction will account for both the magnitude and direction of the momentum change.

Watch out

A common mistake is to think that because the speeds are the same (5 m/s before and after), the momentum change is zero. But momentum is a vector—direction matters! The ball has completely reversed its velocity, so the momentum change is definitely non-zero.

Let me work through this systematically:

  1. Convert the mass to SI units The mass is given as 150 g, which we need in kilograms:

m=150 g=0.15 kgm = 150 \text{ g} = 0.15 \text{ kg}

  1. Identify the initial and final velocities

    Initial velocity: u=(3i^+4j^)\mathbf{u} = (3\hat{i} + 4\hat{j}) m/s

    Final velocity: v=−(3i^+4j^)\mathbf{v} = -(3\hat{i} + 4\hat{j}) m/s

    Notice that v=−u\mathbf{v} = -\mathbf{u}. The ball has reversed direction completely.

  2. Calculate the velocity change

v−u=−(3i^+4j^)−(3i^+4j^)\mathbf{v} - \mathbf{u} = -(3\hat{i} + 4\hat{j}) - (3\hat{i} + 4\hat{j})

=−3i^−4j^−3i^−4j^= -3\hat{i} - 4\hat{j} - 3\hat{i} - 4\hat{j}

=−6i^−8j^ m/s= -6\hat{i} - 8\hat{j} \text{ m/s}

  1. Find the change in momentum Multiply the velocity change by the mass:

Δp=m(v−u)=0.15×(−6i^−8j^)\Delta \mathbf{p} = m(\mathbf{v} - \mathbf{u}) = 0.15 \times (-6\hat{i} - 8\hat{j})

=−0.9i^−1.2j^ kg m/s= -0.9\hat{i} - 1.2\hat{j} \text{ kg m/s}

This can be written as −(0.9i^+1.2j^)-(0.9\hat{i} + 1.2\hat{j}) kg m s−1^{-1}.

Tip

When a ball bounces or reverses direction elastically (same speed, opposite direction), the momentum change is always Δp=−2mu\Delta \mathbf{p} = -2m\mathbf{u}. Here: −2×0.15×(3i^+4j^)=−(0.9i^+1.2j^)-2 \times 0.15 \times (3\hat{i} + 4\hat{j}) = -(0.9\hat{i} + 1.2\hat{j}).

✓Final answer

The correct option is (C) −(0.9i^+1.2j^)-(0.9\hat{i} + 1.2\hat{j}) kg m s−1^{-1}.

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