Question

An 8 kilogram mass is travelling north at 4 m/s and west at 3 m/s. The 8 kilogram mass has a completely inelastic collision with a 2 kilogram mass. Their final velocity is straight north at the same speed that the 8 kilogram mass initially. Find the speed and angle of the 2 kilogram mass before the collision.

Answer 1

let m1 = 8 kg, v1x = -3 m/s, v1y = 4 m/s
m2 = 2 kg,
let v2 is the velocity of m2 just before the collsion,

after the collsion, v = sqrt(4^2 + 3^2)

= 5 m/s
let East be +x axis

Apply conservation of momentum in x direction

m1*v1x + m2*v2x = (m1 + m2)*vx

8*(-3) + 2*v2x = 0

v2x = 8*3/2

= 16 m/s

Apply conservation of momentum in y-direction

m1*v1y + m2*v2y = (m1 + m2)*vy

8*4 + 2*v2y = (8 + 2)*5

v2y = (50 - 32)/2

= 9 m/s

so, initial speed of m2, v2 = sqrt(v2x^2 + v2y^2)

= sqrt(16^2 + 9^2)

= 18.4 m/s <<<<<<<<-------Answer

angle of v2 : theta = tan^-1(v2y/v2x)

= tan^-1(9/16)

= 29.4 degrees with North of East <<<<<<<<-------Answer