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Question

# A ball with initial velocity 1 m/s strikes a stationary ball of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two balls after collision (assuming collinear collision), is

A
12 m/s
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B
2 m/s
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C
12 m/s
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D
14 m/s
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Solution

## The correct option is B √2 m/sLet m be the mass of each ball. Applying momentum conservation, P1i+P2i=P1f+P2f mv0+(m×0)=mv1+mv2 ⇒v0=v1+v2 ............(1) Let the original kinetic energy be K0 From the data given in the question, K1f+K2f=1.5K0 Applying Kinetic energy conservation, K1i+K2i=1.5 K0 ⇒12mv21+12mv22=32(12mv20) ⇒v212+v222=34v20 ⇒v21+v22=32v20 ........(2) From (1), by squaring on both sides, we get v20=(v1+v2)2 ⇒v21+v22+2v1v2=v20 ⇒2v1v2=v20−(v21+v22) Substituting (2) in the above equation, 2v1v2=v20−32v20 ⇒2v1v2=−12v20 So, (v1−v2)2=v21+v22−2v1v2 =32v20−(−12v20)=2v20 ⇒v1−v2=√2v20=√2v0=√2×1=√2 m/s [∵v0=1 m/s] Hence, option (b) is the correct answer.

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