What is the final velocity of the lighter object? (See below)

Two objects whose masses are 1 kg and 4 kg collide with equal and opposite velocities of magnitude 10 m^-1. They bounce off one another, and the heavier object has a final velocity of 5ms^-1 in its original direction. What is the final velocity of the lighter object?

2 Answers
May 14, 2018

The problem is incorrectly specified. The heavier object continues on at -2" m/s" and the lighter object rebounds at -22" m/s"

Explanation:

Given: m_1 = 1" kg", m_2 = 4" kg", vec(u_1)= 10" m/s", vec(u_2) = -10" m/s", vec(v_2) = -5" m/s"

NOTE: I have chosen the direction of the velocity of the lighter object to be positive and the direction of the velocity of the heavier object to be negative.

Take the following equation from this Momentum reference to check whether the initial information is consistent:

vec(v_2) = (m_2-m_1)/(m_1+m_2)vec(u_2)+ (2m_1)/(m_1+m_2)vec(u_1)

We know all of these values:

vec(v_2) = (4" kg"-1" kg")/(1" kg"+4" kg")(-10" m/s")+ (2(1" kg"))/(1" kg"+4" kg")(10" m/s")

vec(v_2) = -2" m/s" color(red)(larr "This is inconsistent"

Let's find the velocity of m_1:

vec(v_1) = (m_1-m_2)/(m_1+m_2)vec(u_1)+ (2m_2)/(m_1+m_2)vec(u_2)

vec(v_1) = (1" kg"-4" kg")/(1" kg"+4" kg")(10" m/s")+ (2(4" kg"))/(1" kg"+4" kg")(-10" m/s")

vec(v_1) = -22" m/s"

May 14, 2018

10 m/s in the direction opposite its original motion.

Explanation:

"Momentum before = momentum after"

m_1*u_1 + m_2*u_2 = m_1*v_1 + m_2*v_2

1 "kg"*(-10 m/s) + 4 "kg"*10 m/s = 1 "kg"*v_1 + 4 "kg"*5 m/s

Solving for v_1

v_1 = (1 "kg"*(-10 m/s) + 4 "kg"*10 m/s - 4 "kg"*5 m/s)/(1 kg)

v_1 = (-10 "kg"*m/s + 40 "kg"*m/s - 20 "kg"m/s)/(1 kg)

v_1 = (10 cancel(kg)*m/s)/(1 cancel(kg)) = 10 m/s

Since v_1 is positive but u_1 was negative, this is the opposite direction of its original motion.

I hope this helps,
Steve