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Q103P

Expert-verifiedFound in: Page 255

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**In Fig. 9-80, block 1 of mass **${{\mathit{m}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{6}}{\mathbf{.}}{\mathbf{6}}{\mathbf{}}{\mathit{k}}{\mathit{g}}$** is at rest on a long frictionless table that is up against a wall. Block 2 of mass **${{\mathit{m}}}_{{\mathbf{2}}}$** is placed between block 1 and the wall and sent sliding to the left, toward block 1, with constant speed ${{\mathit{v}}}_{\mathbf{2}\mathbf{i}}$** **. Find the value of **${{\mathit{m}}}_{{\mathbf{2}}}$** 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 .

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

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

**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.**

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

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 ${m}_{2}=2.2kg$ .

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