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Question

A ball of mass 1kg dropped from 9.8m height strikes the ground and rebounds to a height of 4.9m. If the time of contact between ball and ground is 0.1s, then find impulse and average force acting on the ball.


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Solution

Step 1: Given Data

Mass of the ball m=1kg

The initial velocity of the ball u=0

Let the final velocity when it just strikes the ground be v1.

Let the initial velocity for rebound be v2.

Let the first height be h1=9.8m

Let the second height after collision be h2=4.9m

Let the acceleration due to gravity be g=9.8m/s2

Change in time t=0.1s

Step 2: Calculate final velocity when it just strikes the ground

For the first case, according to Newton's third law of motion

v2=u2+2aS

v12=0+2gh1

v12=2×9.8×9.8

v1=9.82m/s

Step 3: Calculate initial velocity for rebound

For the rebound case, the final velocity will get zero, and acceleration due to gravity will be negative for upward motion.

According to Newton's third law of motion

v2=u2+2aS

0=v2-2gh2

v22=2×9.8×4.9

v2=2×2×4.9×4.9

v2=2×4.9=9.8m/s

Step 4: Calculate the Impulse

Therefore the initial moment is

pi=-mv1j^

Therefore the final moment is

p2=mv2j^

We know that impulse is equal to a change in momentum.

p=pf-pi

=mv2j^--mv1j^

=mv2+v1j^

=19.82+9.8j^

=9.82+1j^=9.8×2.4j^=23.52j^Ns

Step 5: Calculate the average force

We know that impulse can be given as,

I=Fdt=p

For average force, we can write

Favg=pt

=23.520.1=235.2N

Hence, the impulse is 23.52Ns in the positive y direction, and the average force acting on the ball is 235.2N.


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