Ncert Exercises & Solutions

A cricket ball of mass 150 g has an initial velocity \(\small {u = \left(3 \hat{i} + 4 \hat{j} \right) \text {ms}^{- 1}}\) and a final velocity \(\small {v = - \left( 3 \hat{i} + 4 \hat{j} \right) \text{ms}^{- 1}}\), after being hit. The change in momentum (final momentum-initial momentum) is (in kgm/s)
1. \(\text {zero}\)
2. \(-\left ( 0.45\hat{i}+0.6\hat{j} \right ) \)
3. \(-\left ( 0.9\hat{i}+1.2\hat{j} \right ) \)
4. \(-5\left ( \hat{i} +\hat{j}\right ) \)

Step 1: Find the change in momentum of the ball.

\begin{array}{ll} Given, & u = (3 \hat{i} + 4 \hat{j}) m / s \\ and & v = - (3 \hat{i} + 4 \hat{j}) m / s \end{array}

Mass of the ball = 150 g = 0.15 kg

∆

p = Change in momentum

\begin{array}{l} = Final momentum - Initial momentum \\ = mv - mu \\ = m (v - u) = (0 .15) [- (3 \hat{i} + 4 \hat{j}) - (3 \hat{i} + 4 \hat{j}]] \\ = (0 .15) [- 6 \hat{i} - 8 \hat{j}] \\ = - [0 .15 \times 6 \hat{i} + 0 .15 \times 8 \hat{j}] \\ = - [0 .9 \hat{i} + 1 .20 \hat{j}] \\ Hence, Δp = - [0 .9 \hat{i} + 1 .2 \hat{j}] \end{array}

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