Question

object a with a mass of 1.5 kilograms is moving with a velocity of +11.2 meters/second (moving in the +x direction). it has a perfectly elastic collision with stationary object b that has a mass of 4.5 kilograms. after the collision, object b travels with a velocity of +5.6 meters/second. what is the final velocity for object a?

a. +3.8 meters/second in the same direction

b. -3.8 meters/second in the opposite direction

c. +5.6 meters/second in the same direction

d. -5.6 meters/second in the opposite direction

e. -6.2 meters/second in the opposite direction

Explanation:

Step1: Recall Elastic Collision Formulas

In a perfectly elastic collision, both momentum and kinetic energy are conserved. The formula for conservation of momentum is \( m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f} \), and for conservation of kinetic energy (which gives the relative velocity formula) \( v_{1i}-v_{2i}=-(v_{1f}-v_{2f}) \). Since object B is stationary initially, \( v_{2i} = 0 \). So the relative velocity formula simplifies to \( v_{1i}=-(v_{1f}-v_{2f}) \), or \( v_{1f}=v_{2f}-v_{1i} \).

Step2: Substitute Values

Given \( m_1 = 1.5 \, \text{kg} \), \( v_{1i}=11.2 \, \text{m/s} \), \( m_2 = 4.5 \, \text{kg} \), \( v_{2f}=5.6 \, \text{m/s} \). Using the relative velocity formula: \( v_{1f}=5.6 - 11.2=-5.6 \, \text{m/s} \)? Wait, no, let's check with momentum conservation.

Momentum conservation: \( m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f} \)

Substitute \( v_{2i}=0 \): \( 1.5\times11.2 + 4.5\times0=1.5v_{1f}+4.5\times5.6 \)

Calculate left side: \( 1.5\times11.2 = 16.8 \)

Right side: \( 1.5v_{1f}+25.2 \)

Set equal: \( 16.8 = 1.5v_{1f}+25.2 \)

Subtract 25.2: \( 16.8 - 25.2=1.5v_{1f} \)

\( -8.4 = 1.5v_{1f} \)

Divide by 1.5: \( v_{1f}=\frac{-8.4}{1.5}=-5.6 \, \text{m/s} \)? Wait, no, that's not matching. Wait, maybe I made a mistake. Wait, the relative velocity formula: \( v_{1i}-v_{2i}=v_{2f}-v_{1f} \) (since \( v_{1i}-v_{2i}=-(v_{1f}-v_{2f}) \) rearranges to \( v_{1i}-v_{2i}=v_{2f}-v_{1f} \)). So \( 11.2 - 0 = 5.6 - v_{1f} \)

Then \( 11.2=5.6 - v_{1f} \)

Subtract 5.6: \( 11.2 - 5.6=-v_{1f} \)

\( 5.6=-v_{1f} \)

So \( v_{1f}=-5.6 \, \text{m/s} \). Wait, but let's check momentum again. Wait, no, wait the momentum calculation:

Left side: \( 1.5\times11.2 = 16.8 \)

Right side: \( 1.5v_{1f}+4.5\times5.6=1.5v_{1f}+25.2 \)

So \( 16.8=1.5v_{1f}+25.2 \)

\( 1.5v_{1f}=16.8 - 25.2=-8.4 \)

\( v_{1f}=\frac{-8.4}{1.5}=-5.6 \, \text{m/s} \). Wait, but let's check the options. Option D is -5.6 m/s in the opposite direction. Wait, but let's re-express the relative velocity. Wait, maybe I messed up the relative velocity formula. The correct relative velocity formula for elastic collision is \( v_{1i} - v_{2i} = v_{2f} - v_{1f} \). So with \( v_{2i}=0 \), \( v_{1i}=v_{2f}-v_{1f} \), so \( v_{1f}=v_{2f}-v_{1i}=5.6 - 11.2=-5.6 \, \text{m/s} \). So the final velocity of A is -5.6 m/s, which is in the opposite direction (since the initial was +x, negative is -x direction).

Answer:

D. -5.6 meters/second in the opposite direction