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

First, we need to find the potential energy at point A: 
U = mgh = (3.0 kg)(9.81 m/s^2)(0.4 m) = 11.8 J 
This potential energy is transformed into kinetic energy as the mass slides down the track, so: 
K = U = 11.8 J 
The kinetic energy can be expressed in terms of the speed v as: 
K = (1/2)mv^2 
So: 
(1/2)(3.0 kg)v^2 = 11.8 J 
v^2 = (2(11.8 J))/(3.0 kg) 
v = 1.7 m/s

Conservation of momentum can be used to find the final speed of the system after the collision: 
m1v1 + m2v2 = (m1 + m2)vf 
(3.0 kg)(1.7 m/s) + (1.4 kg)(0 m/s) = (3.0 kg + 1.4 kg)vf 
vf = (3.0 kg)(1.7 m/s) / 4.4 kg 
vf = 1.2 m/s

Therefore, the speed of the masses after the collision is 1.9 m/s.

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