A ball with a mass of # 3 kg# is rolling at #16 m/s# and elastically collides with a resting ball with a mass of # 1 kg#. What are the post-collision velocities of the balls?

Question

A ball with a mass of # 3  kg# is rolling at #16   m/s# and elastically collides with a resting ball with a mass of #   1    kg#. What are the post-collision velocities of the balls?

Lily Adams · Accepted Answer

#vec v_1^'=24-16=8" "m/s#

#vec v_2^'=24" "m/s# #vec P_1=m_1*vec v_1" momentum of " m_1" before collision" # #vec P_2=m_2*v_2" momentum of " m_2 " before collision"# #vec Sigma P_b=vec P_1+vec P_2" total momentum before collision"# #Sigma vec P_b=m_1*vec v_1+m_2*vec v_2# #Sigma vec P_b=3.16+1*0=48 " "kg*m/s# #"post collision velocities :"# #vec v_1^'=(2*Sigma vec P_b)/(m_1+m_2)-vec v_1# #vec v_1^'=(2*48)/(3+1)-16# #vec v_1^'=96/4-16# #vec v_1^'=24-16=8" "m/s# #vec v_2^'=(2*Sigma vec P_b)/(m_1+m_2)-vec v_2# #vec v_2^'=(2*48)/(3+1)-0# #vec v_2^'=96/4# #vec v_2^'=24" "m/s#

Axel Acevedo · Answer

undefined To find the post-collision velocities of the balls, you can use the principle of conservation of momentum and the equation for elastic collisions. The post-collision velocities can be calculated using the following formulas:

$$ v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_{1i} + \frac{2m_2}{m_1 + m_2} \cdot v_{2i} $$
$$ v_{2f} = \frac{2m_1}{m_1 + m_2} \cdot v_{1i} + \frac{m_2 - m_1}{m_1 + m_2} \cdot v_{2i} $$

Where:
- $ v_{1f} $ and $ v_{2f} $ are the final velocities of the two balls,
- $ v_{1i} $ is the initial velocity of the first ball (3 kg),
- $ v_{2i} $ is the initial velocity of the second ball (1 kg),
- $ m_1 $ is the mass of the first ball (3 kg),
- $ m_2 $ is the mass of the second ball (1 kg).

Plugging in the given values:

$$ v_{1f} = \frac{3 - 1}{3 + 1} \cdot 16 + \frac{2 \cdot 1}{3 + 1} \cdot 0 $$
$$ v_{2f} = \frac{2 \cdot 3}{3 + 1} \cdot 16 + \frac{1 - 3}{3 + 1} \cdot 0 $$

$$ v_{1f} = \frac{2}{4} \cdot 16 + \frac{2}{4} \cdot 0 $$
$$ v_{2f} = \frac{6}{4} \cdot 16 + \frac{-2}{4} \cdot 0 $$

$$ v_{1f} = 8 + 0 $$
$$ v_{2f} = 12 - 0 $$

Therefore, the final velocity of the first ball ($ v_{1f} $) is 8 m/s, and the final velocity of the second ball ($ v_{2f} $) is 12 m/s.

Leo Abeyta · Answer

undefined To find the post-collision velocities of the balls:

1. Use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
2. Calculate the initial momentum of the system using the formula: initial momentum = mass × velocity.
3. Since the collision is elastic, kinetic energy is conserved as well. Use the formula for kinetic energy: KE = 0.5 × mass × velocity^2.
4. Set up equations based on conservation of momentum and conservation of kinetic energy.
5. Solve the system of equations to find the post-collision velocities of the balls.

Given:
- Mass of ball 1 (before collision): $m_1 = 3 \text{ kg}$
- Mass of ball 2 (before collision): $m_2 = 1 \text{ kg}$
- Velocity of ball 1 (before collision): $v_1 = 16 \text{ m/s}$
- Velocity of ball 2 (before collision): $v_2 = 0 \text{ m/s}$

Using conservation of momentum: $m_1v_1 + m_2v_2 = m_1v_{1f} + m_2v_{2f}$

Using conservation of kinetic energy: $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Solving these equations:

From conservation of momentum: $3(16) + 1(0) = 3v_{1f} + 1v_{2f}$
$48 = 3v_{1f} + v_{2f}$ ...(1)

From conservation of kinetic energy: $\frac{1}{2}(3)(16)^2 + \frac{1}{2}(1)(0)^2 = \frac{1}{2}(3)(v_{1f})^2 + \frac{1}{2}(1)(v_{2f})^2$
$192 = 24v_{1f} + \frac{1}{2}v_{2f}^2$ ...(2)

Using equation (1) to express $v_{2f}$ in terms of $v_{1f}$:
$v_{2f} = 48 - 3v_{1f}$

Substitute this expression for $v_{2f}$ into equation (2) and solve for $v_{1f}$:

$192 = 24v_{1f} + \frac{1}{2}(48 - 3v_{1f})^2$

Solve for $v_{1f}$ and then find $v_{2f}$ using equation (1).