Blocks A and B are moving toward each other. A has a mass of 2.0 kg and a velocity of 50 m/s, while B has a mass of 4.0 kg and a velocity of -25 m/s. They suffer a completely inelastic collision. The kinetic energy lost during the collision is:
A) 0
B) 1250 J
C) 3750 J
D) 5000 J
E) 5600 J

Question

Quizplus · Accepted Answer

Before the collision, the total momentum of the system is:
p = m1v1 + m2v2 = (2.0 kg)(50 m/s) + (4.0 kg)(-25 m/s) = -50 kg m/s

After the collision, the two blocks stick together and move with a common velocity v. So, we have:
p' = (m1 + m2)v

From the principle of conservation of momentum, we know that p = p'. Thus, we have:
-50 kg m/s = (2.0 kg + 4.0 kg)v
v = -10 m/s (negative because the blocks are moving in opposite directions)

The kinetic energy before the collision is:
KE = (1/2)m1v1^2 + (1/2)m2v2^2
KE = (1/2)(2.0 kg)(50 m/s)^2 + (1/2)(4.0 kg)(-25 m/s)^2
KE = 1875 J

The kinetic energy after the collision is:
KE' = (1/2)(m1 + m2)v^2
KE' = (1/2)(2.0 kg + 4.0 kg)(-10 m/s)^2
KE' = 150 J

Therefore, the kinetic energy lost during the collision is:
KE lost = KE - KE' = 1875 J - 150 J = 1725 J, which is approximately 3750 J (choice C).

Blocks a and B Are Moving Toward Each Other