A rubber ball of mass 50 g falls from height of 1m and rebounced to a height of 0.5m. Find the impulse and the force b/w the ball and the ground if the time for which they are in contact was 0.1 sec.????
Answers
★Given:-
- Mass of ball = 50g
- Height = 1m
- Height rebouned = 0.5m
- Time for which they were in contact = 0.1s
★To find:-
- Impulse and the force b/w the ball and the ground.
★Solution:-
From the question,
→Initial velocity of ball = u = 0
→Final velocity of ball = v
→Distance = s = 1m
Using the formula,
✦v²- u²= 2as
Putting values,
⇒ v²- 0² = 2(10)(1)
⇒ v² = √20 m/s
Velocity of rebound of the ball :-
Here,
→S = 0.5 m
→Final velocity= 0
Putting values in the formula,
⇒v² - u²= 2as
⇒0² - (u)² = 2(-10)0.5
⇒u = -√10 m/s
We know:
✦Impulse = Mass×change in velocity
⇒ (50× 10^-3)× (√20 - (-√10)
⇒ (0.05)×(√10(√2 + 1))
⇒ (0.05 × 2.414√10)kg m/s
⇒ 0.38 kg m/s
Using,
✦Average force = Impulse/Time
= 0.38/0.1
= 3.8 N
Hence,
The impulse and the force b/w the ball and the ground are 0.38kg m/s & 3.8N respectively.
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Calculate the impulse and the average force between the ball ... if the time during which they are in contact was 0.1 sec. 1.