Two blocks move along a linear path on a nearly frictionless air track. One block, of mass 0.105 kg, initially moves to the right at a speed of 4.50 m/s, while the second block, of mass 0.210 kg, is initially to the left of the first block and moving to the right at 7.10 m/s. Find the final velocities of the blocks, assuming the collision is elastic.
velocity of the 0.105 kg block | = | |
velocity of the 0.210 kg block | = |
Suppose final speed of 0.105k block is v1 to the left and speed of 0.210kg block is v2 to the right .
Using elastic collision equation,
Velocity of seperation = Velocity of Approach
v1 + v2 = 4.50 + 7.10
v2 = 11.6 - v1 1
Now Applying momentum conservation,
0.105 x 4.50 - 0.210 x 7.10 = 0.210v2 - 0.105v1
-1.02 = 0.210 (11.6 - v1 ) - 0.105v1
v1 = 10.97 m/s
v2 = 11.6 - 10.97 = 0.633 m/s
Velocity of 0.105 kg block = 10.97 m/s to the left
Velocity of 0.210 kg block = 0.633 m/s to the right
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