Answer:
Magnitude of impulse, |J| = 4 kg-m/s
Explanation:
It is given that,
Mass of cart 1, 
Mass of cart 2,
Initial speed of cart 1,
Initial speed of cart 2,
(stationary)
The carts stick together. It is the case of inelastic collision. Let V is the combined speed of both carts. The momentum remains conserved.

V = 1 m/s
The magnitude of the impulse exerted by one cart on the other is given by:


J = -4 kg-m/s
or
|J| = 4 kg-m/s
So, the magnitude of the impulse exerted by one cart on the other 4 kg-m/s. Hence, this is required solution.
Answer:
The impact force will be same for both the cases.
Explanation:
The rate of change of momentum is known as the Impulse and is given by:

where
I = Impulse


Now,
In first case both the cars are identical and have same velocity and in the second case, the wall is stationary.
Also, in both the cases the car does not bounces off the things it hit.
Thus

Thus
Impact force, 
Therefore, impact force is same for both the cases.
Answer:

Explanation:
Here we know that the glider is accelerated uniformly from rest to final speed of 25.7 m/s in total distance of d = 46.9 m
so we will have


d = 46.9
so for uniformly accelerated motion we have



now we will find the total work done given as change in kinetic energy



now power is given as



Answer:
Explanation:
delta V = v * alpha * delta T
= V * 0.00053 * (92.2 - 55.0)
= 0.019716 V
percentage that the owner
= [delta V / V] * 100
= [0.019716 V / V] * 100
= 1.9716 %