A 3.0-kg mass is released from rest at point A of a circular frictionless track of radius 0.40 m as shown in the figure.The mass slides down the track and collides with a 1.4-kg mass that is initially at rest on a horizontal frictionless surface.If the masses stick together,what is their speed after the collision?
A)2.1 m/s
B)1.7 m/s
C)1.9 m/s
D)1.5 m/s
E)2.3 m/s

Question

A 3.0-kg mass is released from rest at point A of a circular frictionless track of radius 0.40 m as shown in the figure.The mass slides down the track and collides with a 1.4-kg mass that is initially at rest on a horizontal frictionless surface.If the masses stick together,what is their speed after the collision? 
A)2.1 m/s 
B)1.7 m/s 
C)1.9 m/s 
D)1.5 m/s 
E)2.3 m/s

Quizplus · Accepted Answer

The potential energy at point A equals the kinetic energy at point B (since there is no friction). So, mgh = (1/2)mv^2. Solving for v at point B, we get v = sqrt(2gh) = sqrt(2*9.8*0.4) = 1.98 m/s. 
After the collision, we can use conservation of momentum: 
(m1 + m2) * Vf = m1Vi1 + m2Vi2 (where Vf is the final velocity of both masses stuck together, and Vi1 and Vi2 are the initial velocities of the two separate masses)
Since Vi2 = 0 (the second mass is at rest initially), we can simplify the equation to:
(3.0 kg + 1.4 kg) * Vf = 3.0 kg * 1.98 m/s 
Solving for Vf, we get Vf = 1.90 m/s, which is closest to answer choice C.

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