Q57P In Fig. 9-61, a ball of mass m =... [FREE SOLUTION]

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Chapter 9: Q57P (page 251)

In Fig. 9-61, a ball of mass m = 60 gis shot with speed $v_{i} = 22 m s$ into the barrel of a spring gun of mass M = 240 ginitially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. Assume that the increase in thermal energy due to friction between the ball and the barrel is negligible. (a) What is the speed of the spring gun after the ball stops in the barrel? (b)What fraction of the initial kinetic energy of the ball is stored in the spring?

Short Answer

Expert verified

The final speed of the spring gun is 4.4 m/s
Fraction of initial K.E. of the ball stored in the spring is 0.80

Step by step solution

Step 1: Given Data

Mass of the ball, $m_{b} = 60 g = 0.06 k g$

Initial speed of ball, $m_{i b} = 22 m / s$

Mass of the spring gun, $m_{s} = 240 g = 0.24 k g$

Initial speed of spring gun, $v_{i s} = 0 m / s$

Determining the concept

Use the principle of conservation of momentum, initial and final kinetic energy of collision to findthefinal speed of spring gunandfraction of initial K.E. stored in the spring.According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formulae are as follow:

$P_{i} = P_{f}$

$P = m$

$K = \frac{1}{2} m v^{2}$

where, m is mass, v is velocity, P is linear momentum and K is kinetic energy.

(a) Determining the final speed of spring gun

To findfinal speed of spring gunapplyingthe principle of conservation of momentum,

Total momentum $\overset{鈫�}{P_{i}}$ before collision = Total momentum after collision $\overset{鈫�}{P_{f}}$

For the given situation,

Total initial momentum = Initial momentum of ball + Initial momentum of spring gun.

$\overset{鈫�}{P_{i}} = \overset{鈫�}{P_{i b}} + \overset{鈫�}{P_{i s}}$

As initially spring gun is at rest, $\overset{鈫�}{P_{i s}} = 0$ ,

$\overset{鈫�}{P_{i}} = \overset{鈫�}{P_{i b}} = m_{b} v_{i b} . . . . . . . . . (1)$

Total final momentum = final momentum of ball + final momentum of spring gun

As after collision,theball sticks in the barrel, final speed of both ball and the spring gun will bethe same.

${\overset{鈫�}{P}}_{f} = m_{b} v_{f b} + m_{s} v_{f s} v_{f b} = v_{f s} = v_{f} {\overset{鈫�}{P}}_{f} = (m_{b} + m_{s}) v_{f} . . . . . . . . . . (2)$

Equating equation (1) and (2),

$m_{b} v_{i b} = (m_{b} + m_{s}) v_{f} v_{f} = \frac{m_{s} v_{i b}}{(m_{b} + m_{s})} v_{f} = \frac{0.06 脳 22}{0.06 + 0.24} v_{f} = 4.4 m / s$

Hence, the final speed of the spring gun is 4.4 m/s.

(b) Determining the fraction of initial K.E. of the ball stored in the spring

Now, to find out initial and final K.E. energy of the system,

$\begin{array}{l} K_{i} = \frac{1}{2} m_{b} v_{i b}^{2} \\ K_{i} = \frac{1}{2} 脳 0.06 脳 22^{2} \\ K_{i} = 14.25 J . . . . . . . (3) \end{array}$

$\begin{array}{l} K_{f} = \frac{1}{2} (m_{b} + m_{s}) v_{f}^{2} \\ K_{f} = \frac{1}{2} 脳 (0.06 + 0.24) 脳 4 . 4^{4} \\ K_{f} = 2.904 鈥� J . . . . . . (4) \end{array}$

Taking ratio of,

$\begin{array}{l} \frac{K_{f}}{K_{i}} = \frac{2.904}{14.52} \\ \frac{K_{f}}{K_{i}} = \frac{1}{5} \end{array}$ .

Fraction of energy stored in the spring $= 1 - \frac{1}{5}$

Fraction of energy stored in the spring $= \frac{4}{5}$

Fraction of energy stored in the spring $= 0.80$

Hence, the fraction of initial K.E. of the ball stored in the spring is 0.80.

Therefore, the principle of conservation of momentum, initial and final kinetic energy of collision can be used to find the final speed of the spring ball $v_{f}$ and fraction of initial kinetic energy stored in the spring.