A mass M with an initial momentum pM = 3000 kg.m.s-1 in a direction 67° from the horizontal collides in a perfectly inelastic manner with another mass N, whose momentum is pN = 2000 kg.m.s-1 in a direction 45° from the horizontal, as shown in the diagram on the right. Non elastic oblique collision

Calculate (i) the angle of motion, θ, of the combined bodies, and (ii), the magnitude,|pc| of the momentum of the combined bodies after collision.

Answer:

Non elastic oblique collision (i) Resolve the initial momentum vectors in the x-direction:
pMx = pMcos67° = 3000 x 0.3907 = 1172 kg.m.s-1,
and pNx = pNcos45° = 2000 x 0.7071 = 1414 kg.m.s-1.
In the same way, resolve the initial momentum vectors in the y-direction:
pMy = pMsin67° = 3000 x 0.9205 = 2762 kg.m.s-1,
and pNy = pNsin45° = 2000 x 0.7071 = 1414 kg.m.s-1.
Then, the final momentum in the x-direction, pCx = pMx + pNx = 1172 + 1414 = 2586 kg.m.s-1
and the final momentum in the y-direction, pCy = pMy + pNy = 2762 + 1414 = 4176 kg.m.s-1.
Now, tanθ = pCy/pCyx = 4176/2586 = 1.615, hence θ = 58.2°.

(ii) The magnitude of pC = |pC| = (pCx2 + pCy2)½ = (25862 + 41762)½ = 4912 kg.m.s-1