A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the same road at 14.0 m/s . The two vehicles remain locked together after the collision.

a. What is the velocity (magnitude and direction) of the two vehicles just after the collision?
b. At what speed should the truck have been moving so that both it and the car are stopped in the collision?
c. Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

(A) V = 9.89m/s

(B) U = -2.50m/s

(C) ΔK.E = –377047J

(D) ΔK.E = –257750J

Explanation:

The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.

This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.

See attachment below for full solution.

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.