A scooter of 40 kg mass moving with velocity $4 m / s$ collides with another scooter of 60 kg mass and moving with velocity $2 m / s$ . After collision the two scooters stick to each other the loss in kinetic energy-

392 J

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440 J

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48 J

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110 J

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Solution

The correct option is D 48 J
loss in KE $= \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (u_{1} - u_{2})^{2}$
$= \frac{1}{2} \frac{40 \times 60}{100} (4 - 2)^{2} = 48 J$

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A scooter of 40 kg mass moving with velocity 4m/s collides with another scooter of 60 kg mass and moving with velocity 2m/s. After collision the two scooters stick to each other the loss in kinetic energy-

The correct option is D 48 Jloss in KE=12m1m2m1+m2(u1−u2)2=1240×60100(4−2)2=48J

A scooter of 40 kg mass moving with velocity $4 m / s$ collides with another scooter of 60 kg mass and moving with velocity $2 m / s$ . After collision the two scooters stick to each other the loss in kinetic energy-

The correct option is D 48 J
loss in KE $= \frac{1}{2} \frac{m_{1} m_{2}}{m_{1} + m_{2}} (u_{1} - u_{2})^{2}$
$= \frac{1}{2} \frac{40 \times 60}{100} (4 - 2)^{2} = 48 J$