A ball with a mass of #12 kg# moving at #3 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 #12 kg# moving at #3 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?

Charles Aeschliman · Accepted Answer

#vec v_1^'=-9/11 " "m/s#
#v_2^'=24/11 " "m/s# #"Momentums before collision :"# #vec P_1=m_1*vec v_1" (for the first object)"# #vec P_1=12*3=" "36 kg.m/s# #vec P_2=m_2*vec v_2" (for the second object)"# #vec P_2=21*0=0# #Sigma vec P_b=vec P_1+ vec P_2" (total momentum before collision)"# #Sigma vec P_b=36+0=36" " kg*m/s# #"Momentums after collision :"# #vec P_1^'=m_1*v_1^'" (for the first object)"# #vec P_1^'=12*v_1^'# #vec P_2^'=m_2*v_2^'" (for the second object)"# #vec P_2^'=21*v_2^'# #Sigma vec P_a=vec P_1^'+vec P_2^'" (total momentum after collision)"# #Sigma vec P_a=12*v_1^'+21*v_2^'# #Sigma vec P_b=Sigma vec P_a" (conservation of momentum")# #36=12*vec v_1^'+21*vec v_2^'" (1)"# #"We can fallow two ways to solve problem:"# #"1 )............................................................"# #v_1+v_1^'=v_2+v_2^' " (using the conservation of kinetic energy)"# #3+v_1^'=0+v_2^'# #v_2^'=3+v_1^' " (2)"# #"let' use (1)" # #36=12*vec v_1^'+21*(3+vec v_1^')# #36=12*vec v_1^'+63+21*vec v_1^'# #36-63=33*vec v_1^'# #-27=33*vec vec v_1^'# #vec v_1^'=-27/33" "vec v_1^'=-9/11 " "m/s# #"now ,let's use (2)"# #vec v_2^'=3+vec v_1^'# #vec v_2^'=3-9/11" "vec v_2^'=(33-9)/11" "v_2^'=24/11 " "m/s# #"2)............................................................."# #v_1^'=(2*Sigma vec P_b)/(m_1+m_2)-v_1# #v_1^'=(2*36)/(12+21)-3# #v_1^'=72/33-3" "v_1^'=(72-99)/33 " "v_1^'=-27/33=-9/11 " m/s" # #v_2^'=(2*Sigma vec P_b)/(m_1+m_2)-v_2# #v_2^'=(2*36)/(12+21)-0# #v_2^'=72/33=" "v_2^'=24/11 " "m/s#

Nicholas Agrusa · Answer

undefined To solve for the final velocity of the second ball, we can use the principle of conservation of momentum:

$$m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$$

Given:
- $m_1 = 12 \, \text{kg}$ (mass of the first ball)
- $v_1 = 3 \, \text{m/s}$ (initial velocity of the first ball)
- $m_2 = 21 \, \text{kg}$ (mass of the second ball)
- $v_2 = ?$ (initial velocity of the second ball, assumed to be zero)
- $v_f = 0$ (final velocity of both balls)

Using the conservation of momentum equation:

$$12 \, \text{kg} \times 3 \, \text{m/s} + 21 \, \text{kg} \times 0 = (12 \, \text{kg} + 21 \, \text{kg}) \times 0$$

$$36 \, \text{kg}\cdot\text{m/s} = 33 \, \text{kg}\cdot\text{m/s}$$

$$v_2 = \frac{36 \, \text{kg}\cdot\text{m/s}}{21 \, \text{kg}}$$

$$v_2 ≈ 1.71 \, \text{m/s}$$

To find the kinetic energy lost as heat in the collision, we can use the principle of conservation of mechanical energy:

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

Given:
- $KE_{\text{initial}} = \frac{1}{2}m_1v_1^2$ (initial kinetic energy of the first ball)
- $KE_{\text{final}} = \frac{1}{2}(m_1 + m_2)v_f^2$ (final kinetic energy of both balls, assumed to be zero)
- $m_1 = 12 \, \text{kg}$ (mass of the first ball)
- $v_1 = 3 \, \text{m/s}$ (initial velocity of the first ball)
- $m_2 = 21 \, \text{kg}$ (mass of the second ball)
- $v_f = 0$ (final velocity of both balls)

Using the kinetic energy equations:

$$KE_{\text{initial}} = \frac{1}{2}(12 \, \text{kg})(3 \, \text{m/s})^2$$
$$KE_{\text{initial}} = 54 \, \text{J}$$

$$KE_{\text{final}} = \frac{1}{2}(12 \, \text{kg} + 21 \, \text{kg}) \times 0^2$$
$$KE_{\text{final}} = 0$$

$$\text{heat energy lost} = 54 \, \text{J} - 0$$
$$\text{heat energy lost} = 54 \, \text{J}$$

Therefore, the second ball is moving at approximately $1.71 \, \text{m/s}$, and $54 \, \text{J}$ of kinetic energy was lost as heat in the collision.