5. A 360 gram cart and a heavier 480 gram cart are initially moving at a speed of 0.45 m/s tothe right at the bottom of a hill as shown. They are suddenly pushed apart by releasing acompressed spring on the larger cart. The smaller cart ends up at the top of the 60.0 cm tallhill on the left with a speed of 1.20m/s. (Friction and air resistance can be neglected.)How fast will the heavier cart be moving at the top of a 15 cm hill on the right?1.20 m/sv= ?0.45 m/s

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Description: 5. A 360 gram cart and a heavier 480 gram cart are initially moving at a speed of 0.45 m/s tothe right at the bottom of a hill as shown. They are suddenly pushed apart by releasing acompressed spring on the larger cart. The smaller cart ends up at t

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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5. A 360 gram cart and a heavier 480 gram cart are initially moving at a speed of 0.45 m/s to
the right at the bottom of a hill as shown. They are suddenly pushed apart by releasing a
compressed spring on the larger cart. The smaller cart ends up at the top of the 60.0 cm tall
hill on the left with a speed of 1.20m/s. (Friction and air resistance can be neglected.)
How fast will the heavier cart be moving at the top of a 15 cm hill on the right?
1.20 m/s
v= ?
0.45 m/s

Expert Solution

Step 1

Scenario: Two carts of different masses initially moving right are suddenly pushed away from each other due to a spring. After this, both carts go in the opposite direction.

Given :

Lighter gram cart weight $\to m_{1} = 360 g m = 0.360 k g$

Heavier gram cart weight $\to m_{2} = 480 g m = 0.480 k g$

Initial velocity of both carts $\to u = 0.45 m / s$

Final velocity of both cart’s at an instant they pushed apart,

Lighter gram cart's final velocity $\to v^{'} = - 1.20 m / s$

Heavier gram cart's final velocity $\to v$

Lighter gram cart's height $\to h_{1} = 60 c m = 0.6 m$

Heavier gram cart's height $\to h_{2} = 15 c m = 0.15 m$

Step 2

Case 1: Sudden Apart

From momentum conservation :

$(m_{1} + m_{2}) u = m_{1} (- v_{1}') + m_{2} v_{2} (0.360 + 0.480) \times 0.45 = - 0.360 \times v_{1}' + 0.480 \times v_{2}' 0.38 = - 0.360 \times v_{1}' + 0.480 \times v_{2} 38 = - 36 v_{1}' + 48 \times v_{2}' 19 = - 18 \times v_{1}' + 24 \times v_{2}' . . . . . . . . . . . . . . . . (1)$

Step 3

Case 2: Applying energy conservation on the lighter cart;

T.E_i=T.E_f

$\frac{1}{2} m_{1} (- {v_{1}}^{2}) = \frac{1}{2} m_{1} {(- v')}^{2} + m g h m_{1} (\frac{1}{2} \times {v_{1}}^{2}) = m_{1} (\frac{1}{2} \times {(1.20)}^{2} + 9.8 \times 0.6) {(v_{1})}^{2} = 2 \times (\frac{1}{2} \times {(1.20)}^{2} + 9.8 \times 0.6) {(v_{1})}^{2} = {(1.20)}^{2} + 2 \times 9.8 \times 0.6 = 13.2 v_{1}' = \sqrt{13.2} v_{1}' = 3.63 \frac{m}{s}$

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