Two blocks of masses m 1 = 1.75 kg and m 2 = 3.50 kg are each released from rest

Question

Two blocks of masses

m1 = 1.75 kg

and

m2 = 3.50 kg

are each released from rest at a height of

h = 5.60 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.

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

(c) Determine the maximum heights to which

m1

and

m2

rise after the collision.

y1f =

y2f =

v1i =

v2i =

Explanation / Answer

According to the given problem,

Since there is no friction, velocity is independent of the mass of the object.
Both has a velocity v = 2gh = 2*9.81*5.6 = 10.482 m/s
------------------------
Since there is no friction, the collision is elastic.
Choosing a reference frame riding with the right side mass,
Velocity of approach = 2*10.482 = 20.964
v1’ = (m1-m2) u1/ (m1+m2) = (1.75-3.5)* 20.964/ (5.25) = -6.988 m/s
v2’ = 20.964-7.46 =13.976m/s
----------------------------------
In the ground reference frame,
The final velocity of 1.75kg is -6.988-10.482 = -17.47m/s
The final velocity of 3.5kg is 13.976-10.482 = 3.494m/2
---------------------------------
Height is found using
h =v^2/ 2g
The height of 1.75kg is 17.47^2/(2*9.81) = 15.55 m
The height of 3.5 kg is 3.494^2/(2*9.81) = 0.622m

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