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College Physics (Urone)
Found in: Page 291
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

A \(5.50\;kg\) bowling ball moving at \(9.00\;{\rm{m/s}}\)collides with a \(0.850\;{\rm{kg}}\)bowling pin, which is scattered at an angle of \(85.{0^ \circ }\)to the initial direction of the bowling ball and with a speed of\(15.0\;{\rm{m/s}}\). (a) Calculate the final velocity (magnitude and direction) of the owling ball. (b) Is the collision elastic? (c) Linear kinetic energy is greater after the collision. Discuss how spin on the ball might be converted to linear kinetic energy in the collision.

  1. The final velocity of the ball is\(8.83\;{\rm{m/s}}\) .
  2. The collision is inelastic.
  3. The spin of the ball results the decrease in linear kinetic energy.
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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Data

The mass of the bowling ball is \({m_1} = 5.50\;kg\)

The mass of the bowling pin is\({m_2} = 0.850\;{\rm{kg}}\).

The speed of the ball is\(v = 9.00\;{\rm{m/s}}\).

The angle of scattering with the initial is\(\theta = 85.0^\circ \).

Step 2: Calculation of final velocity

(a) Using the conservation of momentum along horizontal direction we get,

\(\begin{aligned}{m_1}{v_1}\cos {\theta _1} + {m_2}{v_2}\cos {\theta _2} &= mv\\{v_1} &= \dfrac{{mv - {m_2}{v_2}\cos {\theta _2}}}{{{m_1}\cos {\theta _1}}}\end{aligned}\)

Substituting the values we get,

\(\begin{aligned}{v_1} &= \dfrac{{5.50 \times 9.00 - 0.850 \times 15.0 \times \cos 85.0^\circ }}{{5.50 \times \cos 5^\circ }}\\ &= 8.83\;{\rm{m/s}}\end{aligned}\)

Step 3: Calculation of kinetic energy

(b) The initial kinetic energy is,

\(\begin{aligned}{\left( {KE} \right)_i} &= \dfrac{1}{2} \times 5.50 \times {9^2}\\ &= 223\;{\rm{J}}\end{aligned}\)

The final kinetic energy is,

\(\begin{aligned}{\left( {KE} \right)_f} &= \dfrac{1}{2} \times 5.50 \times {8.83^2} + \dfrac{1}{2} \times 0.850 \times {15.0^2}\\ &= 310\;{\rm{J}}\end{aligned}\)

As the initial and the final kinetic energies are not equal, the collision is inelastic.

Step 4: How the spin of the ball is converted to linear kinetic energy

(c) The initial kinetic energy will be more, but as soon as the ball spins a little extra, it allows the ball to gain more speed. For this reason, there will be a decrease in rotational kinetic energy, which results in a decrease in linear kinetic energy.

Most popular questions for Physics Textbooks

Must the total energy of a system be conserved whenever its momentum is conserved? Explain why or why not.

Two identical pucks collide on an air hockey table. One puck was originally at rest.

(a) If the incoming puck has a speed of \(6.00 m/s \)andscatters to an angle of ,what is the velocity (magnitude anddirection) of the second puck? (You may use the result that \({\theta _1}-{\theta _2} = 90°\) for elastic collisions of objects that have identicalmasses.)

(b) Confirm that the collision is elastic.

Starting with equations\({m_1}{v_1} = {m_1}{v_1}^\prime \cos {\theta _1} + {m_2}{v_2}^\prime \cos {\theta _2}\) and \(0 = {m_1}{v_1}^\prime \sin {\theta _1} + {m_2}{v_2}^\prime \sin {\theta _2}\)for conservation of momentum in the x - and y -directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses, 1/2mv12 = 1/2mv’12 + 1/2mv’22 + mv’1v’2cos (ɵ1 - ɵ2) as discussed in the text.

A \(90.0\;{\rm{kg}}\)ice hockey player hits a \(0.150\;{\rm{kg}}\) puck, giving the puck a velocity of\(45.0\;{\rm{m/s}}\). If both are initially at rest and if the ice is frictionless, how far does the player recoil in the time it takes the puck to reach the goal \(15.0\;{\rm{m}}\) away?

What is an inelastic collision? What is a perfectly inelastic collision?
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