Two objects of masses 100 g and 200 g are moving along the same line and direction with velocities of 2 ms –1 and 1 ms –1 , respectively. They collide and after the collision, the first object moves at a velocity of 1.67 ms –1 . Determine the velocity of the second object.
Assuming that the first object is object A and the second one is object B, it is given that:
Mass of A (m1) = 100g
Mass of B (m2) = 200g
Initial velocity of A (u1) = 2 m/s
Initial velocity of B (u2) = 1 m/s
Final velocity of A (v1) = 1.67 m/s
Final velocity of B (v2) =?
Total initial momentum = Initial momentum of A + initial momentum of B
= m1u1 + m2u2
= (100g) × (2 m/s) + (200g) × (1 m/s) = 400 g.m.sec-1
As per the law of conservation of momentum, the total momentum before collision must be
equal to the total
momentum post collision.
Therefore, m1u1 + m2u2 = m1v1 + m2v2 = 400 g.m.s-1
Solving for v2 = (100g)(1.67 ms-1) + (200g)(v2) = 400 g.m.s-1
Therefore, v2 \( = \frac{400-167}{200}\,m.s^{-1}\)
v2 = 1.165 m/s
Therefore, the velocity of object B after the collision is 1.165 meters per second.
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