Question

If a 1195 kg car, moving at 6 m/s north, hits and sticks to a 3039 kg truck, moving at 4 m/s east, find the magnitude of the final speed.

Hint: here the masses are moving north and east (or in the x & y directions) - so it's a two dimensional problem. Just like all other two dimensional problems we need to separate the x and y motions, treat each separately, and the put the two directions back together. Use the "hit and stick" or inelastic formula

Answer 1

Since there is no external force applied, So momentum of system will remain conserved.

Using Momentum conservation in x-direction

Pix = Pfx

m1*u1x + m2*u2x = (m1 + m2)*Vx

u1 = 6 m/sec towards north = 0 i + 6 j

u1x = 0 m/sec and u1y = 6 m/sec

u2 = 4 m/sec towards east = 4 i + 0 j

u2x = 4 m/sec and u2y = 0 m/sec

Vx = (1195*0 + 3039*4)/(1195 + 3039)

Vx = 2.87 m/sec

Using Momentum conservation in y-direction

Piy = Pfy

m1*u1y + m2*u2y = (m1 + m2)*Vy

Vy = (1195*6 + 3039*0)/(1195 + 3039)

Vy = 1.69 m/sec

So final Velocity of car + truck will be:

V = Vx i + Vy j = 2.87 i + 1.69 j

Speed = |V| = sqrt (2.87^2 + 1.69^2)

|V| = 3.33 m/sec

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