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Chapter 7: Q28P (page 170)

Two billiard balls of equal mass undergo a perfectly elastic head-on collision. If one ball’s initial speed was 2.00 m/s and the other’s was 3.60 m/s in the opposite direction, what will be their speeds and directions after the collision?

Short Answer

Expert verified

After the collision, the speed of the ball with an initial speed of 2.00 m/s is\( - 3.6\;{\rm{m/s}}\)that of the ball with an initial speed of 3.60 m/s is 2 m/s. Both balls reverse their direction after the collision, i.e., opposite to their initial direction.

Step by step solution

01

Given data

The speed of the first billiard ball is\({v_{\rm{A}}} = 2\;{\rm{m/s}}\).

The speed of the second billiard ball is\({v_{\rm{B}}} = - 3.60\;{\rm{m/s}}\).

02

Relationship between the speeds of the billiard balls after the collision 

Consider the movement of ball A is in the positive direction, and the movement of ball B is in the negative direction. Obtain a relationship for velocity to find the final velocity after the collision.

The relation for relative velocity is

\(\begin{array}{c}{v_{\rm{A}}} - {v_{\rm{B}}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\2\;{\rm{m/s}} + 3.60\;{\rm{m/s}} = - \left( {{{v'}_{\rm{A}}} - {{v'}_{\rm{B}}}} \right)\\{{v'}_{\rm{B}}} = \left( {5.60\;{\rm{m/s}} + {{v'}_{\rm{A}}}} \right).\end{array}\)… (i)

Here, \({v'_{\rm{A}}}\) is the final velocity of ball A, and \({v'_{\rm{B}}}\) is the final velocity of ball B.

03

Calculation of the speed of the first billiard ball after the collision

The relation from the momentum equation is

\(\begin{array}{c}m{v_{\rm{A}}} + m{v_{\rm{B}}} = m{{v'}_{\rm{A}}} + m{{v'}_{\rm{B}}}\\{v_{\rm{A}}} + {v_{\rm{B}}} = {{v'}_{\rm{A}}} + {{v'}_{\rm{B}}}.\end{array}\)

Here, m is the mass of the ball.

Substitute the values from equation (i) in the above equation.

\(\begin{array}{c}2\;{\rm{m/s}} + - 3.60\;{\rm{m/s}} = {{v'}_{\rm{A}}} + \left( {5.60\;{\rm{m/s}} + {{v'}_{\rm{A}}}} \right)\\2{{v'}_{\rm{A}}} = - 7.2\;{\rm{m/s}}\\{{v'}_{\rm{A}}} = - 3.6\;{\rm{m/s}}\end{array}\)

04

Calculation of the speed of the second billiard ball after the collision

Substitute the value of speed in equation (i).

\(\begin{array}{l}{{v'}_{\rm{B}}} = \left( {5.60\;{\rm{m/s}} - 3.6\;{\rm{m/s}}} \right)\\{{v'}_{\rm{B}}} = 2\;{\rm{m/s}}\end{array}\)

Thus, \({v'_{\rm{A}}} = - 3.6\;{\rm{m/s}}\) is the final speed of the first billiard ball whose final direction is opposite to its initial direction, and \({v'_{\rm{B}}} = 2\;{\rm{m/s}}\) is the final speed of the second billiard ball whose final direction is also opposite to its initial direction.

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Most popular questions from this chapter

A 0.450-kg hockey puck, moving east with a speed of 5.80 m/s, has a head-on collision with a 0.900-kg puck initially at rest. Assuming a perfectly elastic collision, what will be the speed and direction of each puck after the collision?

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A radioactive nucleus at rest decays into a second nucleus, an electron, and a neutrino. The electron and neutrino are emitted at right angles and have momenta of \(9.6 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\) and \(6.2 \times {10^{ - 23}}\;{\rm{kg}} \cdot {\rm{m/s}}\), respectively. Determine the magnitude and the direction of the momentum of the second (recoiling) nucleus.

Billiard ball A of mass \({m_{\rm{A}}} = 0.120\;{\rm{kg}}\) moving with speed \({v_{\rm{A}}} = 2.80\;{\rm{m/s}}\) strikes ball B, initially at rest, of mass \({m_{\rm{B}}} = 0.140\;{\rm{kg}}\) As a result of the collision, ball A is deflected off at an angle of 30.0° with a speed \({v'_{\rm{A}}} = 2.10\;{\rm{m/s}}\)

(a) Taking the x axis to be the original direction of motion of ball A, write down the equations expressing the conservation of momentum for the components in the x and y directions separately.

(b) Solve these equations for the speed, \({v'_{\rm{B}}}\), and angle, \({\theta '_{\rm{B}}}\), of ball B after the collision. Do not assume the collision is elastic.
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