11. A cricket ball of mass 100 g moving with a speed of
30 m/s is brought to rest by a player in 0.03 s. Find
a. the change in momentum of the ball.
b. the average force applied by the player.
Answers
Answered by
42
Given:-
A player brough a moving ball to rest by applying a force.
- mass of ball, m = 100 g = 0.1 kg
- Initial velocity of cricket ball, u = 30 m/s
- Final velocity of ball, v = 0 m/s
- time taken for change in velocity, t = 0.03 s
To find:-
- Change in the momentum of the ball, Δp =?
- The average force applied by the player, F =?
The formula required:-
- Formula for change in momentum
Δp = ( Final p) - (Initial p) = mv - mu
- Formula to calculate force
F = Δp / t
[ Where p is momentum, m is mass, v is final velocity, u is initial velocity, F is force and t is the time ]
Solution:-
Calculating the change in momentum of the ball
→ Δp = mv - mu
→ Δp = (0.1 × 0) - (0.1 × 30)
→ Δp = -3 kg m/s
Now, calculating the force applied by the player
→ F = Δp / t
→ F = -3 / 0.03
→ F = -100 N
Therefore,
- There is a change of -3 kg m/s in the momentum of the ball. (negative sign represents that momentum of the ball decreased with the time)
- and, A force of magnitude 100 N was applied by the Player against the direction of motion of the ball.
Answered by
2
ANSWER
I guess the question is the rate of change in momentum
Mass=0.1kg
Initial Momentum=m×speed (u)=0.1×30=3
Final Momentum= m×rest (v)=0.1×0=0
Change in Momentum= Final-Initial =0−3=−3
Rate of change in momentum=(change in momentum)÷time taken
=(−3)÷0.03
=−100
I hope it will help you
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