Question

21
9
object s of mass 0.16 kg is moving forward at a speed of 0.50 m s⁻¹. object t of mass 0.10 kg is at rest. object s strikes the object t. after collision, object t moves forward at a speed of 0.50 m s⁻¹. assume no energy is transferred to other objects, what is the speed of object s after collision?

a 0.0 m s⁻¹
b 0.19 m s⁻¹
c 0.31 m s⁻¹
d 0.50 m s⁻¹

Explanation:

Step1: Identify the principle (Conservation of Kinetic Energy)

Since no energy is transferred to other objects, the total kinetic energy before collision equals the total kinetic energy after collision. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \). Let \( m_S = 0.16 \, \text{kg} \), \( v_{S, \text{before}} = 0.50 \, \text{m/s} \), \( m_T = 0.10 \, \text{kg} \), \( v_{T, \text{after}} = 0.50 \, \text{m/s} \), and \( v_{S, \text{after}} \) be the speed of object S after collision.

Step2: Write the conservation of KE equation

\[
\frac{1}{2}m_S v_{S, \text{before}}^2 = \frac{1}{2}m_S v_{S, \text{after}}^2 + \frac{1}{2}m_T v_{T, \text{after}}^2
\]
We can cancel out the \( \frac{1}{2} \) from all terms:
\[
m_S v_{S, \text{before}}^2 = m_S v_{S, \text{after}}^2 + m_T v_{T, \text{after}}^2
\]

Step3: Substitute the known values

Substitute \( m_S = 0.16 \), \( v_{S, \text{before}} = 0.50 \), \( m_T = 0.10 \), and \( v_{T, \text{after}} = 0.50 \) into the equation:
\[
0.16 \times (0.50)^2 = 0.16 \times v_{S, \text{after}}^2 + 0.10 \times (0.50)^2
\]

Step4: Calculate the left - hand side and the right - hand side terms

First, calculate \( (0.50)^2=0.25 \)
Left - hand side: \( 0.16\times0.25 = 0.04 \)
Right - hand side: \( 0.16v_{S, \text{after}}^2+0.10\times0.25=0.16v_{S, \text{after}}^2 + 0.025 \)

Step5: Solve for \( v_{S, \text{after}}^2 \)

Set up the equation:
\[
0.04=0.16v_{S, \text{after}}^2 + 0.025
\]
Subtract \( 0.025 \) from both sides:
\[
0.04 - 0.025=0.16v_{S, \text{after}}^2
\]
\[
0.015 = 0.16v_{S, \text{after}}^2
\]
Then, solve for \( v_{S, \text{after}}^2 \):
\[
v_{S, \text{after}}^2=\frac{0.015}{0.16}=\frac{15}{160}=\frac{3}{32}\approx0.09375
\]

Step6: Solve for \( v_{S, \text{after}} \)

Take the square root of both sides:
\[
v_{S, \text{after}}=\sqrt{0.09375}\approx0.306\approx0.31 \, \text{m/s}
\]

Answer:

\( 0.31 \, \text{m/s} \) (corresponding to option C)