A ball with a mass of 55 kgkg is rolling at 44 ms^-1ms1 and elastically collides with a resting ball with a mass of 33 kgkg. 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=3.88v1=3.88 ms^-1ms1 and v_2=0.12v2=0.12 ms^-1ms1.

Explanation:

Momentum before the collision: p=mv=5*4=20p=mv=54=20 kgms^-1kgms1

Kinetic energy before the collision: E_k=1/2mv^2=1/2xx5xx4^2=40Ek=12mv2=12×5×42=40 JJ

(the stationary ball has a momentum of 00 kgms^-1kgms1 and a kinetic energy of 00 JJ before the collision)

The fact that momentum and kinetic energy are conserved mean that the total momentum after the collision will be 2020 kgms^-1kgms1 and the total kinetic energy after the collision will be 4040 JJ.

Call the 55 kgkg ball '1' and the 33 kgkg ball '2'.

For momentum:

m_1v_1+m_2v_2=20m1v1+m2v2=20 - call this equation (A)

For kinetic energy:

1/2m_1v_1^2+1/2m_2v_2^2=4012m1v21+12m2v22=40 - call this equation (B)

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

5v_1+3v_2=205v1+3v2=20 - (A')

5/2v_1^2+3/2v_2^2=4052v21+32v22=40 - (B')

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

5v_1^2+3v_2^2=805v21+3v22=80 - (B'')

Rearranging equation A', v_1=4-3/5v_2v1=435v2

Substituting this value of v_1v1 into B'':

5(4-3/5v_2)^2+3v_2^2=805(435v2)2+3v22=80

Expanding the square:

5(16-24/5v_2+9/25v_2^2)+3v_2^2=805(16245v2+925v22)+3v22=80

80-24v_2+9/5v_2^2+3v_2^2=808024v2+95v22+3v22=80

Subtract 80 from both sides and rearrange:

24/5v_2^2=24v_2245v22=24v2

Multiplying through by 5 (I hate fractions):

24v_2^2=120v_224v22=120v2

Dividing through by v_2v2:

24v_2=12024v2=120

v_2=0.2v2=0.2 ms^-1ms1

Substituting back into A':

v_1=4-3/5(0.2)=3.88v1=435(0.2)=3.88 ms^-1ms1

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

Jul 5, 2016

"please look over the animation for details"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