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Standard XII

Physics

Conservation of Momentum

A truck of ma...

Question

A truck of mass 15 tons moving with $1$ ms $^{- 1}$ collides with a stationary truck of mass 10 tons and automatically connected to move together. The common velocity is

1

^{- 1}

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0

^{- 1}

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0.4

^{- 1}

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0.6

^{- 1}

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Solution

The correct option is D $0.6$ ms $^{- 1}$
Using momentum conservation,

$m_{1} u_{2} + m_{2} u_{2} = V (m_{1} + m_{2})$

$15 \times 1 + 10 \times 0 = V (15 + 10)$

$\Rightarrow v = \frac{15}{25} = 0.6$ ms $^{- 1}$ .

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A truck of mass 15 tons moving with 1 ms−1 collides with a stationary truck of mass 10 tons and automatically connected to move together. The common velocity is

The correct option is D 0.6 ms−1Using momentum conservation,m1u2+m2u2=V(m1+m2)15×1+10×0=V(15+10)⇒v=1525=0.6ms−1.

A truck of mass 15 tons moving with $1$ ms $^{- 1}$ collides with a stationary truck of mass 10 tons and automatically connected to move together. The common velocity is

The correct option is D $0.6$ ms $^{- 1}$
Using momentum conservation,

$m_{1} u_{2} + m_{2} u_{2} = V (m_{1} + m_{2})$

$15 \times 1 + 10 \times 0 = V (15 + 10)$

$\Rightarrow v = \frac{15}{25} = 0.6$ ms $^{- 1}$ .