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Expert-verified Found in: Page 255 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # In Fig. 9-80, block 1 of mass ${{\mathbit{m}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{6}}{\mathbf{}}{\mathbit{k}}{\mathbit{g}}$ is at rest on a long frictionless table that is up against a wall. Block 2 of mass ${{\mathbit{m}}}_{{\mathbf{2}}}$ is placed between block 1 and the wall and sent sliding to the left, toward block 1, with constant speed ${{\mathbit{v}}}_{\mathbf{2}\mathbf{i}}$ . Find the value of ${{\mathbit{m}}}_{{\mathbf{2}}}$ for which both blocks move with the same velocity after block 2 has collided once with block 1 and once with the wall. Assume all collisions are elastic (the collision with the wall does not change the speed of block 2). The value of mass ${m}_{2}$ is 2.2 kg .

See the step by step solution

## Step 1: Understanding the given information

i) Mass of block 1, ${m}_{1}is6.6kg$ .

ii) The speed of block 2 is ${v}_{2i}$ .

## Step 2: Concept and formula used in the given question

You use the concept of conservation of momentum to find the mass. You use the equations given in the book 9-75 and 9-76 and can solve for the mass of block 2.

${V}_{1f}=\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}{v}_{2i}\phantom{\rule{0ex}{0ex}}{V}_{2f}=\frac{\left({m}_{2}-{m}_{1}\right)}{{m}_{1}+{m}_{2}}{v}_{2i}$

## Step 3: Calculation for the value of m2 for which both blocks move with the same velocity after block 2 has collided once with block 1 and once with the wall

We can use the below equations to find mass as,

${V}_{1f}=\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}{v}_{2i}\phantom{\rule{0ex}{0ex}}{V}_{2f}=\frac{\left({m}_{2}-{m}_{1}\right)}{{m}_{1}+{m}_{2}}{v}_{2i}$

Here, ${v}_{2i}$ is the initial velocity of block -2.

The velocity of block 2, after bouncing off the wall, will be the same as before but with a negative sign; we can write it as,

${V}_{1f}=-{V}_{2f}$

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

$\frac{2{m}_{2}}{{m}_{1}+{m}_{2}}{v}_{2i}=\frac{\left({m}_{2}-{m}_{1}\right)}{{m}_{1}+{m}_{2}}{v}_{2i}\phantom{\rule{0ex}{0ex}}2{m}_{2}=-{m}_{2}-{m}_{1}\phantom{\rule{0ex}{0ex}}3{m}_{2}={m}_{1}\phantom{\rule{0ex}{0ex}}{m}_{2}=\frac{{m}_{1}}{3}$

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

${m}_{2}=\frac{6.6}{3}\phantom{\rule{0ex}{0ex}}=2.2kg$

Thus, the mass of block 2 is ${m}_{2}=2.2kg$ . ### Want to see more solutions like these? 