Question

A ball of mass 200 gm. rolls with velocity of 10 m/s. It is hit with a bat in the direction of its motion. The velocity changes to 20 m/sec. If the bat is in contact with ball for 0.02 sec, the impulsive force on it will be____ *

Answer 1

Answer:

F=ma or m*(v-u)/t

F= 200/1000( 20-10)/0.02

=20(10)/2

=100N

now,

I= F*t

= 100*0.02

= 5000Ns

Answer 2

Answer :

Given -

• Mass = 200 gm = 200/1000 = 1/10 = 0.2 kg
• Initial velocity = 10 m/s
• Final velocity = 20 m/s
• Time period of the force = 0.02 sec

To Find -

• Impulsive force ?

Solution -

So, F = (mv - mu)/t

⇒ F = (0.2*20 - 0.2*10)/0.02

⇒ F = (4 - 2)/0.02

⇒ F = 2/0.02

⇒ F = 100 N

Now, Impulse = Force * Time

⇒ Impulse = 100 * 0.02

⇒ Impulse = 2 N s

Hence, the impulse force applied on the ball by the bat is 2 N s.