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Standard XII

Physics

Momentum

A clay ball o...

Question

A clay ball of mass m and speed v strikes another metal ball of same mass m, which is at rest. They stick together after collision. The kinetic energy of the system after collision is

\frac{m v^{2}}{2}

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\frac{m v^{2}}{4}

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2 m v^{2}

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m v^{2}

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Solution

The correct option is A $\frac{m v^{2}}{4}$
Applying the momentum conversation:

$m v = 2 m v_{1} \to v_{1} = \frac{v}{2}$

$K E = \frac{1}{2} (2 m) (\frac{v}{2})^{2}$ = $\frac{m v^{2}}{4}$

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A clay ball of mass m and speed v strikes another metal ball of same mass m, which is at rest. They stick together after collision. The kinetic energy of the system after collision is

The correct option is A mv24Applying the momentum conversation:mv=2mv1→v1=v2KE=12(2m)(v2)2=mv24

The correct option is A $\frac{m v^{2}}{4}$
Applying the momentum conversation:

$m v = 2 m v_{1} \to v_{1} = \frac{v}{2}$

$K E = \frac{1}{2} (2 m) (\frac{v}{2})^{2}$ = $\frac{m v^{2}}{4}$