A ball with a mass of #15 kg# moving at #6 m/s# hits a still ball with a mass of #21 kg#. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

Question

A ball with a mass of #15 kg# moving at #6 m/s# hits a still ball with a mass of #21 kg#.  If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

Layla Ahmed · Accepted Answer

#4.3ms^-1# and #75.9J#

Conservation of momentum states that before and after a physical reaction there must always be the same amount of momentum.

Momentum is the product of mass and velocity, or

#p = mv#

where #p# is momentum, #m# is mass and #v# is velocity.

Before the reaction, there is a total momentum of

#15kg * 6ms^-1 + 21kg * 0ms^-1 = 90kgms^-1#

because the question describes the #21kg# mass as stationary, therefore it has a velocity of #0# and a momentum of #0#.

After the reaction, we know that the #15kg# mass stops moving, so the velocity and momentum are #0#. The total momentum also must be #90kgms^-1#, according to the law of conservation of momentum.

Therefore, after the reaction,

#p = 15kg * 0ms^-1 + 21kg * v = 90kgms^-1# #21kg * v = 90kgms^-1# #v = 90/21 ms^-1 = 4.3ms^-1#

The answer is rounded to the first decimal place.

For the second part of the question, kinetic energy is given by the formula

#E = 1/2 mv^2#

where #E# is kinetic energy, #m# is mass, and #v# is velocity.

Before the reaction, total kinetic energy would therefore be

#(15 * 6^2)/2 + (21 * 0^2)/2 = 270kgm^2s^-2 = 270J#

After the reaction, we would have

#(15 * 0^2)/2 + (21 * 4.3^2)/2 = 194.1J#

The change in kinetic energy is

#270J - 194.1J = 75.9J# lost as heat.

Feel free to check my working, there's some room for error-carried-forward due to rounding. Hope it's somewhat clear.

Nicholas Agrusa · Answer

undefined To find the velocity of the second ball after the collision, we can use the principle of conservation of momentum:

$$m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot u_1 + m_2 \cdot u_2$$

where:
- $m_1 = 15 \, \text{kg}$ (mass of the first ball)
- $m_2 = 21 \, \text{kg}$ (mass of the second ball)
- $v_1 = 0 \, \text{m/s}$ (final velocity of the first ball)
- $u_1 = 6 \, \text{m/s}$ (initial velocity of the first ball)
- $u_2 = 0 \, \text{m/s}$ (initial velocity of the second ball)
- $v_2$ (final velocity of the second ball)

Solving for $v_2$:
$$m_1 \cdot u_1 + m_2 \cdot u_2 = m_1 \cdot v_1 + m_2 \cdot v_2$$
$$15 \, \text{kg} \times 6 \, \text{m/s} + 21 \, \text{kg} \times 0 \, \text{m/s} = 15 \, \text{kg} \times 0 \, \text{m/s} + 21 \, \text{kg} \times v_2$$
$$90 \, \text{kg} \cdot \text{m/s} = 21 \, \text{kg} \times v_2$$
$$v_2 = \frac{90 \, \text{kg} \cdot \text{m/s}}{21 \, \text{kg}}$$
$$v_2 \approx 4.29 \, \text{m/s}$$

To find the kinetic energy lost as heat in the collision, we can use the equation:

$$KE_{\text{lost}} = KE_{\text{initial}} - KE_{\text{final}}$$

where:
- $KE_{\text{initial}}$ is the initial kinetic energy of the system (before the collision)
- $KE_{\text{final}}$ is the final kinetic energy of the system (after the collision)

$$KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2$$
$$KE_{\text{initial}} = \frac{1}{2} \times 15 \, \text{kg} \times (6 \, \text{m/s})^2 + \frac{1}{2} \times 21 \, \text{kg} \times (0 \, \text{m/s})^2$$
$$KE_{\text{initial}} = \frac{1}{2} \times 15 \, \text{kg} \times 36 \, \text{m}^2/\text{s}^2 + 0$$
$$KE_{\text{initial}} = \frac{1}{2} \times 15 \, \text{kg} \times 36 \, \text{m}^2/\text{s}^2$$
$$KE_{\text{initial}} = 270 \, \text{J}$$

$$KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$$
$$KE_{\text{final}} = \frac{1}{2} \times 15 \, \text{kg} \times (0 \, \text{m/s})^2 + \frac{1}{2} \times 21 \, \text{kg} \times (4.29 \, \text{m/s})^2$$
$$KE_{\text{final}} = 0 + \frac{1}{2} \times 21 \, \text{kg} \times (4.29 \, \text{m/s})^2$$
$$KE_{\text{final}} = 0 + \frac{1}{2} \times 21 \, \text{kg} \times 18.3841 \, \text{m}^2/\text{s}^2$$
$$KE_{\text{final}} = 194.016 \, \text{J}$$

$$KE_{\text{lost}} = KE_{\text{initial}} - KE_{\text{final}}$$
$$KE_{\text{lost}} = 270 \, \text{J} - 194.016 \, \text{J}$$
$$KE_{\text{lost}} \approx 75.984 \, \text{J}$$

Therefore, the velocity of the second ball after the collision is approximately $4.29 \, \text{m/s}$, and the kinetic energy lost as heat in the collision is approximately $75.984 \, \text{J}$.