A block with mass m_1 = 0.550 kg is released from rest on a frictionless track a

Question

A block with mass m_1 = 0.550 kg is released from rest on a frictionless track at a distance h_1 = 2.95 m above the top of a table. It then collides elastically with an object having mass m_2 = 1.10 kg that is initially at rest on the table, as shown in the figure below. (a) Determine the velocities of the two objects just after the collision. (Enter the magnitude of the velocity.) v_1 = m/s v_2 = m/s (b) How high up the track does the 0.550-kg object travel back after the collision? m (c) How far away from the bottom of the table does the 1.10-kg object land, given that height of the table is h_2 = 165 m? m (d) How far away from the bottom of the table does the 0.550-kg object eventually land? m

Explanation / Answer

a) Conserving momentum for m1

Its initial velocity with which it strikes m2 =u =(2gh)^0.5

u = (2*9.8*2.95)^0.5 = 7.604 m/s

Now conserving momentum for collision

m1u = m1v1 + m2v2

0.55*7.604 = 0.55v1 + 1.1v2

4.18 = 0.55v1 + 1.1v2.................(1)

Conserving energy

0.5m1u^2 = 0.5m1v1^2 + 0.5 m2v2^2

0.55*7.604^2 = 0.55v1^2 + 1.1v2^2

31.80 = 0.55v1^2 + 1.1v2^2.................(2)

Using 1 and 2

31.80 = 0.55((4.18-1.1v2)/0.55)^2 + 1.1v2^2

9.62 = 9.61 + 0.6655v2^2 - 5.06v2 + 0.33275v2^2

v2 = 5.06 m/s

v1 = -2.52 m/s

b) conserving energy

v1^2 = 2gh

h = 0.324m

c) Now

R = vo(2h/g)^0.5 = v2(2h2/g)^0.5 = 2.94m

d)

R = vo(2h/g)^0.5 = v1(2h2/g)^0.5 = 1.46m

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