Q54P A completely inelastic collision... [FREE SOLUTION]

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Chapter 9: Q54P (page 251)

A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0 kg, is moving upward at 20 m/sand the other ball, of mass 2.0 kg, is moving downward at 12 m/s. How high do the combined two balls of putty rise above the collision point? (Neglect air drag)

Short Answer

Expert verified

The two combined balls of putty would rise h = 2.6 m above the collision point (h) .

Step by step solution

Step 1: Given Data

Mass of the ball moving downward is, $m_{d} = 2 kg$

Mass of the ball moving upward is, $m_{u} = 3 kg$

Initial speed of the ball moving downward is, $V_{i d} = - 12 m / s$ (downward direction is taken as negative)

Initial speed of the ball moving upward is, $v_{i u} = 20 m / s$

Determining the concept

By applyingtheprinciple of conservation of momentum and usingtheconcept of conversion of kinetic energy to potential energy at maximum height, find rise (h).

According tothe conservation of momentum, momentum of a system is constant if no external forces are acting on the system.

Formula:

$P_{i} = P_{f} P = m v$

At maximum height, $\frac{1}{2} m v^{2} = m g h$

where, m is mass, v is velocity, P is linear momentum, g is an acceleration due to gravity and his height.

Determining the how high do the two combined balls of putty rise above the collision point (h)

Let鈥檚 assume that the upward direction is positive and downward direction is negative.

To calculate rise in the height of the balls, apply the 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 moving downward + Initial momentum of ball moving upward

role="math" localid="1661242020410" $\overset{鈫�}{P_{i}} = \overset{鈫�}{P_{i d}} + \overset{鈫�}{P_{i u}} \overset{鈫�}{P_{i}} = m_{d} v_{i d} + m_{u} v_{i u} . . . . . . . . . (1)$

Total final momentum = Final momentum of ball moving downward +Final momentum of ball moving upward

${\overset{鈫�}{P}}_{f} = m_{d} v_{f d} + m_{u} v_{f u}$

As after collision, two balls stick to each other, their final speeds will be same.

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

Equating equation (1) and (2),

$m_{d} v_{i d} + m_{u} v_{i u} = (m_{d} + m_{u}) v_{f}$

Final speed of the block can be calculated by,

$v_{f} = \frac{m_{d} v_{i d} + m_{u} v_{i u}}{(m_{d} + m_{u})} = \frac{2 脳 (- 12) + (3 脳 20)}{2 + 3} v_{f} = 7.2 m / s$

The final speed of two ballsis 7.2 m/s

To findtherise in the height of the balls, use $v_{f}$

Also, at maximum height,

$\frac{1}{2} m v_{f}^{2} = m g h$

Cancelling mass m and rearranging the equation for h,

$h = \frac{{(v_{f})}^{2}}{2 (g)} h = \frac{{(7.2)}^{2}}{2 (9.8)} = 2.6 m$

As, the height can be considered as positive value, h = 2.6 m

Hence, the two combined balls of putty would rise h = 2.6 m above the collision point.

Therefore, by applying the principle of conservation of momentum and using the law of conservation of mechanical energy, calculate the height of the balls after they stick together.