Two blocks of masses m 1 = 2.00 kg and m 2 = 4.15 kg are each released from rest

Question

Two blocks of massesm1= 2.00 kg andm2=4.15kg are each released from rest at a height ofh=5.50m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.)

(a) Determine the velocity of each block just before the collision.

v1i = m/s

v2i = m/s

(b) Determine the velocity of each block immediately after the collision.

v1f = m/s

v2f = m/s

(c) Determine the maximum heights to which m1 and m2 rise after the collision.

y1f = m

y2f = m

Two blocks of massesm1= 2.00 kg andm2=4.15kg are each released from rest at a height ofh=5.50m on a frictionless track, as shown in the figure below, and undergo an elastic head-on collision. (Let the positive direction point to the right. Indicate the direction with the sign of your answer.) (a) Determine the velocity of each block just before the collision. v1i = m/s v2i = m/s (b) Determine the velocity of each block immediately after the collision. v1f = m/s v2f = m/s (c) Determine the maximum heights to which m1 and m2 rise after the collision. y1f = m y2f = m

Explanation / Answer

(a)Each "falls" from the same height, 5.50 m. At the bottom of the track, each will have the same speed. We can find that speed from the conservation of energy

Ef = Ei
Ef = 0.5 m v2 = m g h = Ei
v2 = 2 g h
v2 = 2 (9.8 m/s2) ( 5.5 m)
v2 = 107.8 m2/s2
v = 10.38 m/s

That is, the 2.0-kg block, m1, moves to the right with
v1 = + 10.38 m/s
while the 4.0-kg block, m2, moves to the left with
v2 = - 10.38 m/s

(b)The total momentum of the two-block system before the collision is
PTot,i = pi1 + p2i
PTot,i = (2.0 kg) (10.38 m/s) + (4.15 kg) ( - 10.38 m/s)
PTot,i = - 22.32 kg m/s

The masses are not the same so the momentum carried by each block is not the same, even tho' their speeds are the same. So the total momentum is not zero!

The two blocks undergo an elastic collision so their final velocities are given by

v1f = [(m1 - m2)/(m1 + m2)] v1i + [ 2 m2/(m1 + m2) ] v2i
v1f = [(2 kg - 4.15 kg)/(2 kg + 4.15 kg)](10.38 m/s) + [ 2(4.15 kg)/(2 kg + 4.15 kg)]( - 10.38 m/s)

v1f = - 17.64 m/s

Mass m1 moves to the left after the collision.

v2f = [2 m1 / (m1 + m2)] v1i + [ (m2 - m1) / (m1 + m2) ] v2i
v2f = [2 (2 kg) / (2 kg + 4.15 kg)] (10.38 m/s) + [ (4.15 kg - 2 kg) / (2 kg + 4.15 kg)] ( - 10.38 m/s)

v2f = 3.12 m/s

Mass m2 moves to the right after the collision.

(c) By conservation of energy

For block 1,

Eend = Eslide
Eend = Uend = m g h1 = 0.5 m v1f2 = Kslide = Eslide
m g h1 = 0.5 m v1f2
h1 = [0.5 v1f2] / g
h1 = [0.5(17.64 m/s)2] / (9.8 m/s2)

h1 = 15.88 m

For block 2,

h2 = [0.5 v2f2] / g

h2 = [0.5 (3.12 m/s)2] / (9.8 m/s2)

h2 = 0.5 m

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