A 19.6-kg curling stone (A) moving at 1.20 m/s [N] strikes another identical stationary stone (B)...

 

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A 19.6-kg curling stone (A) moving at 1.20 m/s [N] strikes another identical stationary stone (B) off centre, and moves off with a velocity of 1.17 m/s [12.0 E of N]. What will be the velocity of stone B after the collision? Ignore frictional and rotational effects. Given m = 19.6 kg VB = 0 m/s mB = 19.6 kg VA = 1.20 m/s [N] VAL 1.17 m/s [12.0 E of N] + before N after N W S VB = 0 m/s + -E VA = 1.20 m/s Required final velocity of stone B (VB) VA = 1.17 m/s VB 12.0 + W E m+ S Figure 9.48 A 19.6-kg curling stone (A) moving at 1.20 m/s [N] strikes another identical stationary stone (B) off centre, and moves off with a velocity of 1.17 m/s [12.0 E of N]. What will be the velocity of stone B after the collision? Ignore frictional and rotational effects. Given m = 19.6 kg VB = 0 m/s mB = 19.6 kg VA = 1.20 m/s [N] VAL 1.17 m/s [12.0 E of N] + before N after N W S VB = 0 m/s + -E VA = 1.20 m/s Required final velocity of stone B (VB) VA = 1.17 m/s VB 12.0 + W E m+ S Figure 9.48

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Lets denote vBf to be the velocity of mass B after collision ... View the full answer