Question

A player catches a ball of 200g moving with a speed of 2o m/s if the time taken to complete the catch is 0.5 sec the force exerted on the player hand is

Answer 1

Given,mass=0.2kg
initial velocity(u)= 20m/s
final velocity(v)=0m/s
impulse= change in momentum= 0.2×20-0.2×0=4 kg m/s

Impulse= force × time

force= impulse/time=4/0.5=8 N

Answer 2

Answer:

The force exerted on the player's hand is 8 N.

Explanation:

It is given that,

Mass of the ball, m = 200 g = 0.2 kg

Initial velocity of the ball, u = 20 m/s

Final velocity of the ball, v = 0 (as it is caught by the player)

Time taken, t = 0.5 s

We need to find the force exerted on the player's hand. It is calculated as :

$F\times t=m(v-u)$

Since, impulse is equal to the change in momentum of the object. So,

$F=\dfrac{m(v-u)}{t}$

$F=\dfrac{0.2\ kg\times (0-20\ m/s)}{0.5\ s}$

F = -8 N

So, the force exerted on the player's hand is 8 N. Hence, this is the required solution.