Question

A 2.0-kg ball moving at 10 m/s makes an off-center collision with a 3.0-kg ball that is initially at rest. After the collision, the 2.0-kg ball is deflected at an angle of 30° from its original direction of motion and the 3.0-kg ball is moving at 4.0 m/s. Find the speed of the 2.0-kg ball and the direction of the 3.0-kg ball after the collision.
Hint: sin²θ + cos²θ = 1.

speed of 2.0-kg ball	m/s
direction of 3.0-kg ball	° from original direction of motion

Answer 1

ELASTIC COLLISION

m1 = 2 kg                       m2 = 3 kg

speeds before collision

u1x = 10 m/s                   u1y = 0

u2x = 0                         u2y = 0

speeds after collision

v1x = v1*cos30            v1y = v1*sin30

v2x = 4*costheta      v2y = -4*sintheta

initial momentum before collision

Pix = m1*u1x + m2*u2x

after collision final momentum

Pfx = m1*v1x + m2*v2x

from moentum conservation

total momentum is conserved

from momentum conservation total momentum is conserved

Pfx = Pix

m1*u1x + m2*u2x = m1*v1x + m2*v2x

(2*10) + (3*0) = (2*v1*cos30) + (3*4*costheta).....(1)

12*costheta = (20 - 1.732*v1)..........(1)

Pfy = Piy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

(2*0) + (3*0) = (2*v1*sin30) - (3*4*sintheta)

12*sintheta = v1.............(2)

1^2 + 2^2

12^2 = (20-(1.732v1))^2+v1^2

v1 = 5.34 m/s <<<<----------answer

from 2

12*sintheta = 5.34

theta = 26.4 degrees <<<<----------answer