Dynamics(Impulse and Momentum) Questions. Please help! Don`t understand!
1. An 80.0 kg man walks from his house to the store. He runs the 2 km distance in 12 min. What is his average momentum?
2. A 75 g tennis ball is hit by a racket and flies through the air at 30 m/s. If the racket was in contact with the tennis ball for 0.2 ms, what was the average force on the tennis ball during the impact?
3. A 6.00 kg toy car accelerates uniformly from rest to a velocity of 12.0m/s west. What is the impulse on the toy car?
4. A 1.2 kg basketball hits the ground with a velocity of 2.0 m/s. If the ball bounces up with a velocity of 0.6 m/s, what is the ball’s change in momentum?
5. An average net force of 1.9 X 105 N up for 9 s acts on the space shuttle with a mass of 7.84 X 104 kg on. What is the final velocity of the rocket?
1 Answer | Add Yours
First these terms, momentum, impulse and force, they all are vectors that means they are quantities with a direction. So when you specify their value you have to mention the direction also.
1) Momentum = Mass x Velocity
In average terms,
Average momentum = Mass x Average velocity
The avergae velocity of the man = 2000 m/ (12*60) sec towards the direction from house to shop
= 2.778 ms-1
Average momentum = 80 kg x 2.778 ms-1 towards the same direction above.
= 222.24 kgms-1
In this problem we have to make some assumptiosn, otherwise the answer is not complete,
We have to assume that tennis ball is stationary when it was hit by the racket. (as in serve of a tennis match)
Force = Rate of momentum change
Rate of momentum change = (Final momentum - Initial momentum)/ time
Lets take direction of ball after it hit the racket as positive.
Rate of moentum change
= (0.075 x 30 - 0.075 x 0) / 0.0002 kgms-2
= 11250 N
= 11.25 kN in the direction of ball movement
c) Impulse = Force x Time = F x dt
or in other words, I = momentum change
I = F x dt = m x a x dt
a x dt is velocity change dv
Therefore I = m x dv
I = 6 kg x (12 -0) ms-1
I = 72 Ns-1 in west direction (Newtons persecond)
Let's take the downward direction as positive. Then the velocity downward is positive while the velocity upwards is negative
Then the momentum change = Final momentum - Intital momentum
= 1.2 x (-0.6) kgms-1 - 1.2x2 kgms-1
= -0.72 - 2.4 kgms-1
= -3.12 kgms-1
The moemntum change is -3.12 kgms-1 downwards or +3.12 kgms-1 upwards.
You have done a typing mistake here. Therefore I am assuming that 1.9 x 105 is actually 1.9 x 10^5 N (ten to the power 5, otherwoise this doesnt make any sense) and 7.84 x 104 as 7.84 x 10^4 respectively.
First we can find the acceleration of the rocket due to the force using F = ma,
1.9 x 10^5 N = 7.84 x 10^4 kg x a
a = 2.423 ms-2
Assuming the rocket was initially stationary (ie V1 = 0)
V = V1 + a*t
V = 0 + 2.423 ms-2 x 9 s
= 21.807 ms-1
The final velocity is 21.807 ms-1
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