(II) Two bullets are fired at the same time with the same kinetic energy. If one bullet has twice the mass of the other, which has the greater speed and by what factor? Which can do the most work?

Short Answer

Expert verified

The lighter bullet has a higher speed, and the speed of the lighter bullet is \(\sqrt 2 \) times the speed of the heavier bullet.

Both bullets do the same amount of work.

Step by step solution

01

Given data and assumptions

Given data:

The kinetic energies of both bullets are the same.

Assumptions:

Let the mass of the first bullet be \({m_1}\) and the mass of the second bullet be \({m_2}\).

Let the speed of the first bullet be \({v_1}\) and the speed of the second bullet be \({v_2}\).

02

Calculation of speed

The lower mass has a greater speed if the kinetic energies of the two particles of different masses are the same because speed varies inversely with mass.

\(\begin{array}{l}K = \frac{1}{2}m{v^2}\\{v^2} \propto \frac{1}{m}{\rm{ }}\left( {k = {\rm{constant}}} \right)\end{array}\)

Let the mass of the first bullet be twice the mass of the second, i.e.,\({m_1} = 2{m_2}\).

According to the question, the kinetic energies of both bullets are the same. Therefore,

\(\begin{array}{c}\frac{1}{2}{m_1}v_1^2 = \frac{1}{2}{m_2}v_2^2\\{\left( {\frac{{{v_1}}}{{{v_2}}}} \right)^2} = \frac{{{m_2}}}{{{m_1}}}\\{\left( {\frac{{{v_1}}}{{{v_2}}}} \right)^2} = \frac{{{m_2}}}{{2{m_2}}}\\{v_2} = \sqrt 2 {v_1}\end{array}\)

Hence, the lighter bullet has a higher speed, and the speed of the lighter bullet is \(\sqrt 2 \) times the speed of the heavier bullet if the heavier mass is double the lighter mass.

03

Calculation of work

You know that work done is equal to the change in kinetic energy.

Since both bullets have the same kinetic energy, they will do an equal amount of work.

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Most popular questions from this chapter

Two blocks of mass\({m_{\rm{A}}}\)and\({m_{\rm{B}}}\), resting on a frictionless table, are connected by a stretched spring and then released (Fig. 7–48). (a) Is there a net external force on the system before release? (b) Determine the ratio of their speeds,\({v_{\rm{A}}}/{v_{\rm{B}}}\)(c) What is the ratio of their kinetic energies? (d) Describe the motion of the CM of this system. Ignore mass of spring.

(II) A roller-coaster car shown in Fig. 6–41 is pulled up to point 1 where it is released from rest. Assuming no friction, calculate the speed at points 2, 3, and 4.

FIGURE 6–41 Problem 40

(III) A uniform rod AB of length 5.0 m and mass \({\bf{M = 3}}{\bf{.8}}\;{\bf{kg}}\) is hinged at A and held in equilibrium by a light cord, as shown in Fig. 9–67. A load \({\bf{W = 22}}\;{\bf{N}}\) hangs from the rod at a distance d so that the tension in the cord is 85 N. (a) Draw a free-body diagram for the rod. (b) Determine the vertical and horizontal forces on the rod exerted by the hinge. (c) Determine d from the appropriate torque equation.


Question:(II) A woman holds a 2.0 m long uniform 10.0 kg pole, as shown in Fig. 9–78. (a) Determine the forces she must exert with each hand (magnitude and direction). To what position should she move her left hand so that neither hand has to exert a force greater than (b) 150 N? (c) 85 N?



(II) Suppose the roller-coaster car in Fig. 6–41 passes point 1 with a speed of 1.30 m/s. If the average force of friction is equal to 0.23 of its weight, with what speed will it reach point 2? The distance traveled is 45.0 m.

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