First slide

Types of collision

Question

A cricket ball of mass 150 g moving with a speed of $126 {kmh}^{- 1}$ 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 remain in contact for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be

Moderate

A

10.5 N
B

21 N
C

$1.05 \times 10^{4} N$
D

$2.1 \times 10^{4} N$

Solution

Here, $u = - v = 126 {kmh}^{- 1} = 126 \times \frac{5}{18} = 35 m / s$
Change in momentum of the ball,
$Δ p = m (v - u) = \frac{150}{1000} (- 35 - 35)$
$= \frac{3}{20} (- 70) = - \frac{21}{2} kg - {ms}^{- 1}$
Now, force, $F = \frac{Δ p}{Δ t} = \frac{- 21 / 2}{0.001} N = - 1.05 \times 10^{4} N$
Here, negative sign shows that direction of force will be opposite to the direction of movement of the ball before
hitting.

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App