Ncert Exercises & Solutions

A cricket ball of mass $150~\text{g}$ moving with a speed of $126~\text{km/h}$ hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remains in contact for $0.001~\text{s}$, the force that the batsman had to apply to hold the bat firmly at its place would be:
1. $10.5~\text{N}$
2. $21~\text{N}$
3. $1.05\times10^{4}~\text{N}$
4. $2.1\times 10^{4}~\text{N}$

Hint: Apply the concept of conservation of momentum.

Step 2: Find the final velocity of the ball.

$Given, m = 150 g = \frac{150}{1000} kg = \frac{3}{20} kg$
$Δt = time of contact = 0 .001 s$
$u = 126 km / h = \frac{126 \times 1000}{60 \times 60} m / s = 35 m / s$
$v = - 126 km / h = - 35 m / s$

Step 2: Find the change in momentum.

Change in the momentum of the ball

$\begin{array}{l} Δp = m (v - u) = \frac{3}{20} (- 35 - 35) kg - m / s \\ = \frac{3}{20} (- 70) = - \frac{21}{2} \end{array}$

Step 3: Find the force.

$\begin{array}{l} We know that force F = \frac{Δp}{Δt} \\ = \frac{- 21 / 2}{0 .001} N = - 1 .05 \times 10^{4} N \end{array}$

Here, — ve sign showed that force will be opposite to the direction of the movement of the ban before hitting.

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