A car of mass 400 kg travelling at 72 kmph crashes a truck of mass 4000 kg and travelling at 9 kmph in the same direction. The car bounces back with a speed of 18 kmph. The speed of the truck after the impact is
9 kmph
18 kmph
27 kmph
36 kmph

9 kmph

18 kmph

27 kmph

36 kmph

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Solution

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Using momentum conservation,

$m_{c a r} u_{c a r} + m_{t r u c k} u_{t r u c k} = m_{c a r} v_{c a r} + m_{t r u c k} v_{t r u c k}$

$400 \times 72 + 4000 \times 9 = - 18 \times 400 + 4000 \times v$

$v = 18 k m p h$

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Using momentum conservation,

$m_{c a r} u_{c a r} + m_{t r u c k} u_{t r u c k} = m_{c a r} v_{c a r} + m_{t r u c k} v_{t r u c k}$

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