A ball with a mass of 5 kg is rolling at 4 ms^-1 and elastically collides with a resting ball with a mass of 3 kg. What are the post-collision velocities of the balls?

2 Answers
Jul 4, 2016

Momentum is always conserved, and the key word 'elastically' tells us that kinetic energy is also conserved. The calculation below shows how we determine the final velocities of the balls, which are v_1=3.88 ms^-1 and v_2=0.12 ms^-1.

Explanation:

Momentum before the collision: p=mv=5*4=20 kgms^-1

Kinetic energy before the collision: E_k=1/2mv^2=1/2xx5xx4^2=40 J

(the stationary ball has a momentum of 0 kgms^-1 and a kinetic energy of 0 J before the collision)

The fact that momentum and kinetic energy are conserved mean that the total momentum after the collision will be 20 kgms^-1 and the total kinetic energy after the collision will be 40 J.

Call the 5 kg ball '1' and the 3 kg ball '2'.

For momentum:

m_1v_1+m_2v_2=20 - call this equation (A)

For kinetic energy:

1/2m_1v_1^2+1/2m_2v_2^2=40 - call this equation (B)

We can substitute in the known masses to make these simpler:

5v_1+3v_2=20 - (A')

5/2v_1^2+3/2v_2^2=40 - (B')

To make it even a little neater, let's multiply equation B' by 2:

5v_1^2+3v_2^2=80 - (B'')

Rearranging equation A', v_1=4-3/5v_2

Substituting this value of v_1 into B'':

5(4-3/5v_2)^2+3v_2^2=80

Expanding the square:

5(16-24/5v_2+9/25v_2^2)+3v_2^2=80

80-24v_2+9/5v_2^2+3v_2^2=80

Subtract 80 from both sides and rearrange:

24/5v_2^2=24v_2

Multiplying through by 5 (I hate fractions):

24v_2^2=120v_2

Dividing through by v_2:

24v_2=120

v_2=0.2 ms^-1

Substituting back into A':

v_1=4-3/5(0.2)=3.88 ms^-1

We can substitute these values back into A and B to check them.

Jul 5, 2016

"please look over the animation for details"
v_r^'=1 " "m/s
v_b^'=5" " m/s

Explanation:

enter image source here

"solution-1:"
m_r:"the red ball's mass"
m_b:"the blue ball's mass"
v_r:"the velocity of red ball before collision"
v_b:"the velocity of blue ball before collision"
v_r^':"the velocity of red ball after collision"
v_b^':"the velocity of blue ball after collision"

"momentums before collision"
P_r=m_r*v_r=5*4=20 kg*m/s" the red ball's momentum"
P_b=m_b*v_b=3*0=0" the blue ball's momentum"
Sigma P=P_r+P_b=20+0=20 kg*m/s" thee total momentum"

v_r^'=(2*Sigma P)/(m_r+m_b)-v_r

v_r^'=(2*20)/(5+3)-4

v_r^'=40/8-4

v_r^'=5-4=1" " m/s

v_b^'=(2*Sigma P)/(m_r+m_b)-v_b

v_b^'=(2*20)/(5+3)-0

v_b^'=40/8-0

v_b^'=5" "m/s

"Solution -2:"

m_r*v_r+m_b*v_b=m_r*v_r^'+m_b*v_b^'

5*4+3*0=5*v_r^'+3*v_b^'

20=5*v_r^'+3*color(red)(v_b^')" (1)"

v_r+v_r^'=v_b+v_b^'

4+v_r^'=0+v_b^'

color(red)(v_b^')=4+v_r^'" (2)"

"(1) " 20=5*v_r^'+3*(4+v_r^')

20=5v_r^'+12+3v_r^'

20-12=8v_r^'

8=8v_r^'" ; "v_r^'=1" "m/s

"use (2)"

v_b^'=4+1=5" "m/s