A Collision?

Question

A projectile of mass 2 kg approaches a stationary target body at 4.9 m/s. The projectile is deflected through an angle of 59.5° and its speed after the collision is 2.8 m/s. What is the magnitude of the momentum of the target body after the collision?

Answer 1

Mp = Mass of projectile = 2kg

V1px = Initial Velocity projectile in x direction= 4.9 m/sec
V1py = Initial Velocity projectile in y direction =0

V2px = Final Velocity projectile in x direction
=2.8m/s*Cos (59.5)
V2py = Final Velocity projectile in y direction
=2.8m/s*Sin (59.5)


Mt = Mass of Target
V1t = Initial Velocity target=0
V2t = Final Velocity target = V2tx +V2ty
V2tx = Final Velocity projectile in x direction
V2ty = Final Velocity projectile in y direction

Conservation of Momentum is:
Mp*V1p +Mt*V1t = Mp*V2p + Mt*V2t

Solve for Momentum in x direction

Mp*V1px +Mt*V1tx=Mp*V2px +MtV2tx

2kg*4.9m/s +Mt*0=2kg*2.8m/s*Cos (59.5)+MtV2tx
9.8kgm/s+0=2.84kgm/s+MV2tx
MtV2tx= 6.96 kgm/s

Solve for Momentum in y direction
Mp*V1py +Mt*V1ty=Mp*V2py +MtV2ty
0+0=2kg*2.8m/s*Sin (59.5)+MtV2ty
0=4.83+MtV2ty
MtV2ty=-4.83kgm/s

Therefore:
MtVt = sqrt(6.96^2 +4.83^2)
= 8.47 kg*m/sec
Direction = inverse tan (-6.96/4.83)
=-34.8 deg