Two blocks of masses, m1 = 2.00 kg and m2 = 4.00 kg, are each released from rest

ID: 2154576 • Letter: T

Question

Two blocks of masses, m1 = 2.00 kg and m2 = 4.00 kg, are each released from rest at a height of 5.00 m on a frictionless track as shown in the figure, and undergo an elastic head-on collision.

Please find out the answer to each question below.(Please show your work for partial credits)
(a) Determine the velocity of each block just before the collision. (2 points)
(b) Determine the velocity of each block immediately after the collision. (3 points)
(c) Determine the maximum heights to which m1 and m2 rise after the collision. (3 points)

Explanation / Answer

To have an ahead on collision the blocks should traverse in opposite directions. The speed gained by each of the blocks in opposite directions under gravity through the frictionless smooth track is given by :

v^2 = 2gs = 2*9.81*5. Or

(i) v = 9.904544412m/s is the speed in opposite directions before impact. m2 has +9.90..m/s anticlocwise and m1 has -9.90..m/s clock wise say.

ii) The velocities of the blocks after collision is given by:

v1f (final velocity of m1 ) = [2m2b+(m1-m2)a]/(m1+m2) where a and b are the initial velocities of the blocks m1 and m2, in this case -v and +v

=[2*4v+(2-4)(-v)]/(2+4) = 10v/6 = 5v/3 = (5*9.904544412m/s)/3 =+16.5076 m/s anti clockwise.

v2f (final velocity v) = [2m1a+(m2-m1)a]/(m1+m2)=2*2(-v)+(4-2)V]/(2+4) = (-4+2)v/6 = -v/3 = -9.904544412m/3 m/s = 3.3015 clockwise.

iii)

The first block with speed 16.5076m/s could move a height h meter, given by 2gh = 16.5076..^2 Or h = 16.5076^2/(2g) = 13.8889 meter of height , where g is the acceleration due to gravity and is assumed 9.81m/s^2

The second block could go as high as 3.3015^2/(2g) = 0.5556 meter of height.