Question

A player catches a ball of 10g moving with a speed of 5m/s. if the time taken to complete the catch is 0.25s, the force exerted on the player's hand is​
a). 8N
b) 4N
c) 2N
d). 0N

Answer 1

Required Answer:-

The force exerted on the player's hand is the rate of change in momentum of the ball. Here, we have got the value of physical quantities:

• Mass of the ball = 0.01 kg
• Initial velocity = 5 m/s
• Final velocity = 0 m/s (It comes to rest).
• Time taken = 0.25 s

We know,

$\boxed{ \rm{ F = \triangle P/t}}$

Where, F is the force exerted, t is the time taken and ∆P is the change in momentum of the ball.

Initial momentum = mv1

= 0.01 kg × 5 m/s

= 0.05 kg m/s

Final momentum = mv2

= 0.01 kg × 0 m/s

= 0 kg m/s

Then,

Force = mv2 - mv1 / t

= 0 - 0.05 / 0.25 N

= - 0.2 N (or simply 0.2 N)

Here negative sign indicates opposite direction.

Answer 2

Answer:

c) 0.2 N

Explanation:

Given that a player catches a ball of 10g moving with a speed of 5m/s. if the time taken to complete the catch is 0.25s.

We need to find out the force exerted on the player's hand.

Now,

Initially the momentum is mv1 and final momentum is mv2.

So,

Initial momentum = mv1 (where m is 10 g = 0.01 kg and v is 5 m/s)

→ 0.01 × 5

→ 0.05

Final momentum = mv2 (where m is 0.01 kg, as it catches the ball means final velocity is 0.)

→ 0.01 × 0

→ 0

Now,

Force = (Change in momentum)/Time

Force = (Final momentum - Initial momentum)/Time

Force = (0 - 0.05)/0.25

Force = - 0.2 N

(Negative sign means force is in opposite direction)

Therefore, the force exerted on the player's hand is -0.2 N