Question

A 1500-kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500-kg van traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming that the vehicles undergo a perfectly inelastic collision (i.e. they stick together).

Answer 1

From conservation of linear momentum, the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.

COLLISION

There are four types of collision

• Elastic collision

• Perfectly elastic collision

• Inelastic collision

• Perfectly inelastic collision

In elastic collision, both momentum and energy are conserved. While in inelastic collision, only momentum is conserved.

From the given question, the following parameters are given.

• $m_{1}$ = 1500kg

• $v_{1}$ = 25 m/s

• $m_{2}$ = 2500 kg

• $v_{2}$ = 20 m/s

Since the collision is inelastic, they will both move with a common velocity after collision.

Horizontal component

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

1500 x 25 = (1500 + 2500) V

37500 = 4000V

V = 37500 / 4000

V = 9.375 m/s

Vertical component

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

2500 x 20 = (1500 + 2500)V

50000 = 4000V

V = 50000 / 4000

V = 12.5 m/s

The net velocity will be the magnitude of the velocity of the wreckage after collision

V = $\sqrt{9.4^{2} + 12.5^{2} }$

V = $\sqrt{244.61}$

V = 15.6 m/s

The direction will be

Tan Ф = 12.5 / 9.4

Ф = $Tan^{-1}$(1.329)

Ф = 53 degrees.

Therefore,  the magnitude of the velocity of the wreckage after collision is 15.6 m/s while its direction is 53 degrees.

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