Consider a system of two balls with masses 2.1kg and 4.4 kg. The lighter ball tr

Question

Consider a system of two balls with masses 2.1kg and 4.4 kg. The lighter ball travels to the east with a speed of 3m/s whereas the heavier ball travels north with a speed of 4m/s until they collide at right angles. The balls stick together at the point of collision. (a) What is the magnitude of the velocity of the balls after the collision? (b) What is the angle north of east of the velocity after the collision? (c) What is the fraction of the initial kinetic energy of the system lost in this collision? Consider a system of two balls with masses 2.1kg and 4.4 kg. The lighter ball travels to the east with a speed of 3m/s whereas the heavier ball travels north with a speed of 4m/s until they collide at right angles. The balls stick together at the point of collision. (a) What is the magnitude of the velocity of the balls after the collision? (b) What is the angle north of east of the velocity after the collision? (c) What is the fraction of the initial kinetic energy of the system lost in this collision? Consider a system of two balls with masses 2.1kg and 4.4 kg. The lighter ball travels to the east with a speed of 3m/s whereas the heavier ball travels north with a speed of 4m/s until they collide at right angles. The balls stick together at the point of collision. (a) What is the magnitude of the velocity of the balls after the collision? (b) What is the angle north of east of the velocity after the collision? (c) What is the fraction of the initial kinetic energy of the system lost in this collision? (a) What is the magnitude of the velocity of the balls after the collision? (b) What is the angle north of east of the velocity after the collision? (c) What is the fraction of the initial kinetic energy of the system lost in this collision?

Explanation / Answer

a] By momentum conservation,

initial momentum = 2.1*3 i + 4.4*4 j = sqrt(6.3^2+17.6^2) = 18.6936 kgm/s

Final momentum = 18.6936

final velocity v = 18.6936/[2.1+4.4] = 2.876 m/s answer

b] angle = arctan(17.6/6.3) = 70.3 degree

c] fraction = [0.5*[2.1+4.4]*2.876^2]/[0.5*2.1*3^2+0.5*4.4*4^2] = 0.602

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