28. A heavy truck and a light car are moving at the same speed (opposite directi

Question

28. A heavy truck and a light car are moving at the same speed (opposite directions) when they collide into one another head on. After the collision, they stick together. Neglect any outside forces (there is no friction with the road.) Which ONE of the following statements is TRUE? A) The total momentum of the system before the collision is zero. B) Mechanical energy is conserved during this collision. C) During the collision, the magnitude of the force of the car on the truck exactly equals the magnitude of the force of the truck on the car. D) Immediately after the collision, the wreck is at rest. E) The collision is elastic. 29. You push a 2.0 kg box along the ground. It moves with a constant speed of 4.0 m/s. If you are exerting a constant horizontal force of 5 N. what mustbe the coefficient offriction, uk? A) 0 B) 2.5 C) 2.0 D) 0.5 E) 0.25 30. A rigid bar of length L has a hinge at the left end. Four forces (A, B, C, or D) can be applied to the bar. Each force has the same magnitude F, but is applied at different locations at different angles, as shown. IThere is no gravity in this problem!) Which force (A, B, C, or D) produces a torque around the hinge with the greatest magnitude? A, B, C, D from figure, or E) Two or more of these forces tie for greatest magnitude of torque.

Explanation / Answer

28.

Since After collision car and truck stuck together, which means collision is inelastic.

In inelastic collision mechanical energy is not conserved.

Initial momentum = Pi = Mt*Vt + Mc*Vc

given that Vt = -Vc

But since Mt is not equal to Mc, which means momentum is not zero initially.

Since initial momentum is not zero, which means final momentum cannot be zero, due to momentum conservation

So wreck is not at rest after collision.

Which means remaining option C is correct.

Reason for option C is Newton's third law, for every action there is equal and opposite reaction.

29.

Since constant speed which means acceleration is zero, So

Fnet = m*a

Fnet = 0

F - Ff = 0

F = Ff

Ff = frictional force = uk*N

Ff = uk*m*g

uk = F/(mg)

uk = 5/(2*9.81) = 0.25

Correct option is E.

30.

torque = rXF = r*F*sin theta

Net torque due to force at A

Ta = F*3*L/4 - F*L/4 = F*L/2 = 0.5*F*L

Net torque due to force at B

Tb = F*L/2 - F*L/2 = 0

Net torque due to force at C

Tc = rXF = r*F*sin theta

Tc = L*F*sin 45 deg = F*L/sqrt 2 = 0.707*F*L

Net torque due to force at D

Td = rXF = r*F*sin theta

Td = L*F*sin 0 deg = 0

So max torque is in case of C

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