Question

A cricket ball of man 150 g mover at a spord
of 12 m/s and after hitting by the bat it is
deflected back at the speed of 20 m ee. If the
bat and the ball remained in contact for 0.02
then, the impulse and average force exerted
on the ball by the bat are respectively.
(1) 24 N5; 480 N (2) 24 N: 120 N
(3) 4.8 N-8 : 240 N (4) 48 N-:430 N​

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the ball (m) = 150 grams = 0.15 Kg.
• Initial velocity (u) = 12 m/s.
• Final velocity (v) = 20 m/s.
• Time of contact (t) = 0.02 seconds.

$\huge{\bold{\underline{Explanation:-}}}$

As the Initial velocity is opposite to the Final velocity,

So, it will be negative,

I.e u = - 12 m/s.

$\rule{300}{1.5}$

As we know,

Change in momentum = Impulse imparted on the body.

Now,

$\large{\bold{ \tt I = m(v - u)}}$

Substituting the values,

$\large{I = 0.15 \times (20 - [-12])}$

$\large{I = 0.15 (20 + 12)}$

$\large{I = 0.15 \times 32}$

$\huge{\boxed{\boxed{ \tt I = 4.8 \: Ns}}}$

So,the Impulse imparted on the body is 4.8 N-s.

$\rule{300}{1.5}$

As we know,

$\large{\bold{ \tt F = \dfrac{ \Delta P}{T}}}$

(From Newton's second law)

Substituting the values,

$\large{F = \dfrac{4.8}{0.02}}$

$\large{F = \dfrac{48 \times 100}{2 \times 10}}$

$\large{F = \dfrac{\cancel{48} \times 10 \cancel{0}}{ \cancel{2} \times \cancel{10}}}$

$\large{F = 24 \times 10}$

$\huge{\boxed{\boxed{ \tt F = 240 \: N}}}$

So, the Force imparted on the body is 240N.

$\rule{300}{1.5}$

So, the impulse and average force exerted on the ball by the bat is 4.8 Ns and 240 N respectively.

Therefore,

Option - (3) is correct.