Two blocks of masses m 1 = 2.00 kg and m 2 = 3.90 kg are each released from rest

Question

Two blocks of masses m1 = 2.00 kg and m2 = 3.90 kg are each released from rest at a height of h = 5.70 m 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   = ?

v2i = ?

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

v2f = ?

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

Explanation / Answer

a) using energy conservation to find the velocities of blocks just before collision,

mgh = mv^2 /2

v =sqrt(2gh) = sqrt(2 x 9.81 x 5.70) = 10.58 m/s

v1i = v2i = v =10.58 m/s

b)
using elastic collision equation,

velocity of approach = velocity of ssepearation

2v = v1f - v2f

v1f - v2f = 21.15 ..... (i)

using momentum conservation,

m1v1 + m2v2 = m1v1f + m2v2f

2 x (-10.58) + (3.90 x 10.58) = 2v1f + 3.90v2f

2v1f + 3.90v2f = 20.10 ......(ii)

from (i) and (ii),

v1f = 17.39 m/s

v2f = - 3.76 m/s

c)using mgh = mv^2 /2 to find height,

h = v^2 / 2g

h1 = 17.39^2 / (2 x 9.81)   =19.16 m

h2 = 3.76^2 / (2 x 9.81) = 0.721 m

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