A block with mass m_1 = 0.450 kg is released from rest on a frictionless track a

Question

A block with mass m_1 = 0.450 kg is released from rest on a frictionless track at a distance h_1 = 2.25 m above the top of a table. It then collides elastically with an object having mass m_2 = 0.900 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 0.450-kg object travel back after the collision? m (c) How far away from the bottom of the table does the 0.900-kg object land, given that the height of the table is h_2 = 1.50 m? m (d) How far away from the bottom of the table does the 0.450-kg object eventually land? m

Explanation / Answer

from conservation of energy

PE = KE

m g h = 1/2 m u1^2

u1 = sqrt (2 g h) = sqrt (2 x 9.8 x 2.25) = 6.64 m/s

from conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2

m1u1 = m1v1 + m2v2

from conservation of KE

KE1 + KE2 = KE1' + KE2'

m1u1^2 + 0 = m1v1^2 + m2v2^2 (1/2 gets cancelled both sides)

solving 1 and 2 simulatneously we get

v1 = (m1-m2)u1/(m1 + m2)

v1 = (0.45 - 0.9) x 6.64/(0.45 + 0.9) = -2.21 m/s

v2 = 2m1u1/(m1 + m2)

v2 = 2 x 0.45 x 6.64/(6.64 + 0.9) = 4.43 m/s

Hence, v1 = -2.21 m/s and v2 = 4.43 m/s

b)Again using conservation of energy

1/2 m v^2 = m g h

h = 1/2 v^2/g = 0.5 x (-2.21)^2/9.8 = 0.25 m

Hence, h = 0.25m

c)time taken will be:

t = sqrt (2 h2/g)

t = sqrt (2 x 1.5/9.8) = 0.55 s

x = v2 t

x = 4.43 x 0.55 = 2.44 m

Hence, x = 2.44 m

d)t = 0.55 s (time will be same as height is same)

x' = v1t = 2.21 x 0.55 = 1.22 m

Hence, x' = 1.22 m

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