Question

A ball with a mass of 0.615 kg is initially at rest. It is struck by a second ball having a mass of 0.380 kg , initially moving with a velocity of 0.260 m/s toward the right along the x axis. After the collision, the 0.380 kg ball has a velocity of 0.230 m/s at an angle of 37.4 ∘ above the x axis in the first quadrant. Both balls move on a frictionless, horizontal surface.

What is the magnitude of the velocity of the 0.615 kg ball after the collision?

What is the direction of the velocity of the 0.615 kg ball after the collision?

What is the change in the total kinetic energy of the two balls as a result of the collision?

Answer 1

m1 = 0.615 kg
m2 = 0.380 kg

​Before Collison -
v1= 0
v2 = 0.260 m/s

After Collison -
v1x = v*cos(θ)?
v1y = v*sin(θ)?

v2x = 0.230 * cos(37.4) m/s
v2y = 0.230 * sin(37.4) m/s

Using Momentum Conservation
Initial Momentum = Final Momentum

For X axis -
m2 * v2 = m1* v1x + m2*v2x
0.380 * 0.260 = 0.615* v*cos(θ) + 0.380 * 0.230 * cos(37.4)
v*cos(θ) = 0.04775 ---------1

For Y axis -
0 = - m1* v1y + m2*v2y
0.380 * 0.230 * sin(37.4) = 0.615* v*sin(θ)
v*sin(θ) = 0.0863 --------------2

2/1
tan(θ) = 0.0863/0.04775
θ = 61.04^o
Direction of the velocity of the 0.615 kg ball after the collision, θ = 61.04^o Clockwise from + x axis in 4th Quadrant.

v = 0.0863 / sin(61.04)
v = 0.0986 m/s
Magnitude of the velocity of the 0.615 kg ball after the collision , v = 0.0986 m/s

Initial Kinetic Energy = 0.5*m*v1^2
Final Kinetic Energy = 0.5*m1v^2 + 0.5*m2*v2^2

Change in total kinetic energy = K.Ein - K.Efi
Change in total kinetic energy = 0.5*0.380*0.260^2  - 0.5*0.380*0.230^2 + 0.5*615*0.0986^2 J
Change in total kinetic energy = 3 J