Question

A baseball player grabs a 200 gram baseball which is moving at speed of 30 m/s. The ball comes to rest in about 1/10 second. The force acting
on the player's hand is (Think of the relationship between change of the momentum and force) ​

Answer 1

M=200 GRAM=0.2 KG

U=0

V=30 M/S

T=0.1 SEC

F=?

F=MA ,FROM THIS EQUATION WE CAN FIND OUT THE VALUE OF FORCE.

WE HAVE MASS(0.2) KG,TO FIND FORCE FIRST WE HAVE TO FIND OUT ACCELERATION.(BY EQUATION'V=U+AT)

                   >30=O+AX0.1

               >   0.1A=30

                >A=300M/S²

NOW,F=MXA

        F=0.2X300

        F=60 NEWTON

AAYA SAMAJ.

Answer 2

PLAYER'S HAND EXPERIENCES 60 N FORCE.

Given:

• Mass of ball = 200 gram = 0.2 kg
• Velocity of ball = 30 m/s
• Time taken to stop = 1/10 sec

To find:

• Force experienced by the hand of the player?

Calculation:

Force experienced by the hand of the player will be equal to the change in momentum of the ball with respect to time .

$F = \dfrac{\Delta P}{\Delta T}$

$\implies F = \dfrac{(mu - mv)}{ T}$

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

$\implies F = \dfrac{ 0.2(30- 0)}{ \frac{1}{10} }$

$\implies F =0.2 \times 30 \times 10$

$\implies F =60 \: N$

So, force experienced by the hand of the player is 60 N.