3 hard balls are placed on a frictionless surface. Ball A (red): mass = 3 kg Bal

Question

3 hard balls are placed on a frictionless surface. Ball A (red): mass = 3 kg Ball B (blue): mass = 2 kg Ball C (green): mass = 1 kg Ball A is then given an initial velocity of 5 m/s towards Balls B and C. Ball A collides elastically with Ball B. a) The speed of Ball A after collision is Blank 1 m/s b) The speed of Ball B after collision is Blank 2 m/s Ball B then collides with and sticks to Ball C. c) The speed of Balls B+C after collision is Blank 3 m/s. d) The change in kinetic energy is Blank 4 J?

Explanation / Answer

A = 3 kg

B = 2 Kg

C = 1 Kg

Initial velocity A = 5 m/s

Since the collision of A with B is elastic, but kinetic energy and momentum is conserved

Let the velocity of A after collision be va and that of B be vb

Conservation of momentum:

3*5 = 3*va + 2*vb

=> 15 = 3va + 2 vb (eq 1)

Conservation of kinetic energy

0.5*3*52 = 0.5 *3*va2 + 0.5*2*vb2

=> 37.5 = 1.5 va2 + vb2 (eq 2)

solving eq 1 and eq 2

va = 1 m/s

vb = 6 m/s

a)  The speed of Ball A after collision = 1 m/s

b) The speed of Ball B after collision = 6 m/s

Now B collides with C and sticks, Let the combined speed by v

By conservation of momentum

2*6 = (2+1)*v

=> v = 4 m/s

c) The speed of Balls B+C after collision = 4m/s

d) Change in KE = 0.5*2*62 - 0.5*(2+1)*42 = 36-24 = 12 J

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