Two bodies move towards each other and collide inelastically. The velocity of the first body before impact is 2 m/s and of the second is 4 m/sec.The common velocity after collision is 1 m/s in the direction of the first body. How many times did the K.E. of the first body exceed that of the second body before collision.

4.25

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3.25

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2.25

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1.25

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Solution

The correct option is D 1.25
Using momentum conservation
$m_{1} \times 2 - 4 m_{2} = (m_{1} + m_{2})$
$\Rightarrow m_{1} = 5 m_{2}$

Ratio of K.E $= \frac{\frac{1}{2} \times m_{1} \times (2)^{2}}{\frac{1}{2} \times m_{2} \times (4)^{2}}$

$= \frac{5 m_{2} \times 4}{m_{2} \times 16}$

$= 1.25$

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The correct option is D 1.25Using momentum conservationm1×2−4m2=(m1+m2)⇒m1=5m2Ratio of K.E=12×m1×(2)212×m2×(4)2=5m2×4m2×16=1.25

The correct option is D 1.25
Using momentum conservation
$m_{1} \times 2 - 4 m_{2} = (m_{1} + m_{2})$
$\Rightarrow m_{1} = 5 m_{2}$

Ratio of K.E $= \frac{\frac{1}{2} \times m_{1} \times (2)^{2}}{\frac{1}{2} \times m_{2} \times (4)^{2}}$

$= \frac{5 m_{2} \times 4}{m_{2} \times 16}$

$= 1.25$