Question

An 80-kg man is skating north and accidentally collides with a 20-kg boy skating east. The two travel together after the collision at 2.5 m/s at an angle of 60o north of east. What was the speed of the boy before the collision?

Answer 1

mass, m1=80 kg, m2=20 kg

law of conservation of momentum,

along x-axis,

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

after collision speed of the both bodies moves with same speed,

m1*u1x+m2*u2x=m1*vx+m2*vx

0+m2*u2x=(m1+m2)*vx

20*u2x=100*v*cos(60)

20*u2x=100*2.5*cos(60)

===> u2x=6.25 m/sec

and

along y-axis,

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

80*u1y+0=80*vy+20*vy

80*u1y=(80+20)vy

80*u1y=(100)*v*sin(60)

====> u1y=2.71 m/sec

speed of the m2 bofore collision is, u1y=2.71 m/sec