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Expert-verified Found in: Page 282 ### Physics For Scientists & Engineers

Book edition 9th Edition
Author(s) Raymond A. Serway, John W. Jewett
Pages 1624 pages
ISBN 9781133947271 # A car of mass m traveling at speed v crashes into the rear of a truck of mass 2m that is at rest and in neutral at an intersection. If the collision is perfectly inelastic, what is the speed of the combined car and truck after the collision? (a) ${\mathbit{v}}$(b)$\frac{\mathbf{v}}{\mathbf{2}}$(c) $\frac{\mathbf{v}}{\mathbf{3}}$ (d) ${\mathbf{2}}{\mathbit{v}}$(e) None of those answers is correct.

The speed of the combined car and truck after the collision is $\frac{v}{3}$, so option (c) is the correct answer.

See the step by step solution

## Step 1: Concept

The total momentum of an isolated system (no external forces) is conserved and doesn’t depend on the nature of the forces between the members of the system:

${\mathbit{\Delta }}{\stackrel{\mathbf{\to }}{\mathbf{P}}}_{\mathbf{t}\mathbf{o}\mathbf{t}}{\mathbf{=}}{\mathbf{0}}$

The system may be isolated in terms of momentum but Non-isolated in terms of energy, as in the case of inelastic collisions.

The linear momentum of a particle $\stackrel{\to }{P}$of mass m moving with a velocity $\stackrel{\to }{v}$ is,

$\stackrel{\to }{P}=m\stackrel{\to }{v}$

Where,

$\stackrel{\to }{P}=$The momentum of the particle

$\stackrel{\to }{v}=$Velocity of particle

role="math" localid="1663741365134" $m=$ Mass of the particle

## Step 2: Final velocity

A car of mass m traveling at speed v crashes into the rear of a truck of mass 2m that is at rest and in neutral at an intersection.

Assuming that the collision was head-on so that, after impact, the wreckage moves in the original direction of the car’s motion, conservation of momentum during the impact is the law we are going to use to solve the equation for final velocity as,

$\begin{array}{c}\left({m}_{c}+{m}_{t}\right){v}_{f}={m}_{c}{v}_{0c}+{m}_{t}{v}_{0t}\\ \left({m}_{c}+{m}_{t}\right){v}_{f}={m}_{c}v+{m}_{t}\left(0\right)\\ {v}_{f}=\left(\frac{{m}_{c}}{{m}_{c}+{m}_{t}}\right)v\end{array}$

Here is ${m}_{c}$the mass of the car, ${m}_{t}$is the mass of the truck.

Substitute the values in the above expression, and we get,

$\begin{array}{c}{v}_{f}=\left(\frac{m}{m+2m}\right)v\\ {v}_{f}=\frac{v}{3}\end{array}$

The speed of the combined car and truck after the collision is $\frac{v}{3}$ , so option (c) is the correct answer.

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