As shown in Figure 1 (below on page 6), an object of mass m1 = 0.20 kg slides al
ID: 2237889 • Letter: A
Question
As shown in Figure 1 (below on page 6), an object of mass m1 = 0.20 kg slides along a track labeled ABC. Point A is a vertical height of H = 2.00 m above the track at B-C. Between points A and B the track is frictionless. Between points B and C the track has a coefficient of friction of ?k = 0.35. At the point A at time t=0, mass m1 has an initial velocity of v10 = 15.00 m/s to the right. It then slides down the track to point B and collides with a second object of mass m2 = 0.10 kg which is initially at rest. The collision is elastic. The track to the right of point B is horizontal and is not frictionless as described above. For reference in part (iv) below, let point B have the coordinates x = 0 and y = 0.What is the velocity of each mass at time t=0 when m1 is at point A and m2 is at Point B?
What is the velocity of each mass at point B just before collision?
What is the velocity of each mass at point B just after collision?
How far along the horizontal track towards point C will mass m2 slide before coming to rest?
Explanation / Answer
Velocity of m1 at t= 0 is 15 m/s and of m2 is 0.
Total energy is 1/2*m1v102 + m1gh = 1/2*m1u1b2
1/2*0.2*152 + 0.2*9.81*2 = 1/2*0.2*u1b2
velocity of m1 at B before collison, u1b = 16.25 m/s
velocity of m2 at B just before collision = 0.
Momentum conservation: m1u1b + 0 = m1v1b + m2v2b
0.2*16.25 = 0.2*v1b + 0.1*v2b...................(1)
Elastic collision: 1/2*m1u1b2 = 1/2*m2v2b2 + 1/2*m1v1b2
1/2*0.2*16.252 = 1/2*0.2*v1b2 + 1/2*0.1*v2b2....................(2)
Solving (1) and (2), v1b = 5.42 m/s and v2b = 21.7 m/s
Hence, velocity of m1 after collision is 5.42 m/s and m2 is 21.7 m/s.
Friction force = m2g = 0.35*0.1*9.81 = 0.343 N
Energy of m2 is = 1/2*m2v2b2 = 1/2*0.1*21.72 = 23.54 J
Distance = 23.54/0.343 = 68.6 m
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