A railway engine of mass $2$ tons moving at a speed of $72 k m h^{- 1}$ collides with a wagon at rest. After collision both hava common velocity of $36 k m h^{- 1}$ . Find the mass of the wagon. ( $1 t o n = 1000 k g$ )

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Solution

Given data:
Mass of railway engine, $m_{e} = 2 t o n s$ $= 2000 k g$
Speed of railway engine, $u_{e} = 72 k m h^{- 1}$ $= 72 \times \frac{5}{18} = 20 m s^{- 1}$
Speed after the collision, $v = 36 k m h^{- 1} = 36 \times \frac{5}{18} = 10 m s^{- 1}$
Finding the mass of the wagon, $m_{w}$ :
By the law of conservation of momentum, linear momentum before collision will be equal to linear momentum after collision.
$m_{e} u_{e} + m_{w} u_{w} = (m_{e} + m_{w}) v$
$(2000 \times 20) + (m_{w} \times 0) = (2000 + m_{w}) 10 40000 = (2000 + m_{w}) 10 4000 = 2000 + m_{w} m_{w} = 4000 - 2000 = 2000 k g = 2 t o n s$
Thus, mass of the wagon, $m_{w} = 2 t o n s$ .

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A railway engine of mass 2 tons moving at a speed of 72kmh-1 collides with a wagon at rest. After collision both hava common velocity of 36kmh-1. Find the mass of the wagon. (1ton=1000kg)

A railway engine of mass $2$ tons moving at a speed of $72 k m h^{- 1}$ collides with a wagon at rest. After collision both hava common velocity of $36 k m h^{- 1}$ . Find the mass of the wagon. ( $1 t o n = 1000 k g$ )