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Found in: Page 255

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# In Fig. 9-78, a 3.2 kg box of running shoes slides on a horizontal frictionless table and collides with a 2.0 kg box of ballet slippers initially at rest on the edge of the table, at height h = 0.40 m The speed of the 3.2 kg box is 3.0 m/s just before the collision. If the two boxes stick together because of packing tape on their sides, what is their kinetic energy just before they strike the floor?

The kinetic energy of the boxes just before they strike the floor, $K{E}_{f}$ is 29 J .

See the step by step solution

## Step 1: Understanding the given information

i) Mass of running shoe box, ${m}_{1}is3.2kg$.

ii) Mass of ballet slipper box, ${m}_{2}is2.0kg$ .

iii) Height of the table, ${h}_{i}is0.40m$ .

iv) Speed of the mass ${m}_{1}$ , ${V}_{i1}is3.0m/s$ .

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

You can use the concept of conservation of momentum and conservation of mechanical energy. Using conservation of momentum, you find the initial velocity of the boxes at the edge of the table. Using conservation of mechanical energy, you can find the final velocity at the ground

${m}_{1}{V}_{i1}+{m}_{2}{V}_{i2}={m}_{1}{V}_{f1}+{m}_{2}{V}_{f2}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{i}^{2}+mg{h}_{1}=\frac{1}{2}m{v}_{f}^{2}+mg{h}_{2}$ ${m}_{1}{V}_{i1}+{m}_{2}{V}_{12}={m}_{1}{V}_{f1}+{m}_{2}{V}_{f2}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{i}^{2}+mg{h}_{1}=\frac{1}{2}m{v}_{f}^{2}+mg{h}_{2}$

## Step 3: Calculation for the kinetic energy when two boxes stick together because of packing tape on their sides

First, we find the velocity of the boxes at the edge of the table; both move together, so they have the same velocity, v; we can say that v can be calculated using momentum conservation law as,

$\left(3.2\right)\left(3.0\right)+\left(20\right)\left(0\right)=\left(3.2\right)v+\left(2.0\right)v\phantom{\rule{0ex}{0ex}}v=\frac{9.6}{5.2}\phantom{\rule{0ex}{0ex}}=1.846m/s$

Now, this is the initial velocity for the next projectile motion. We know that both boxes initially have kinetic energy and potential energy, which will convert to kinetic energy at the ground.

You get,

$K{E}_{i}+P{E}_{i}=K{E}_{f}+P{E}_{f}\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}^{2}+mg{h}_{1}=K{E}_{f}+mg{h}_{f}$

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

$\frac{1}{2}\left(5.2kg\right)\left(1.846m/{s}^{2}\right)+\left(5.2kg\right)\left(9.8m/{s}^{2}\right)\left(0.40m\right)=K{E}_{f}+\left(5.2\right)\left(9.8\right)\left(0\right)\phantom{\rule{0ex}{0ex}}K{E}_{f}=29.24\left(1kg.{m}^{2}/{s}^{2}×\frac{1J}{1kg.{m}^{2}/{s}^{2}}\right)\phantom{\rule{0ex}{0ex}}=29.24J\phantom{\rule{0ex}{0ex}}=29J$

The kinetic energy of the boxes just before they strike the floor, $K{E}_{f}is29J$ .