A ball \"a\" is shot with a speed u=1.4 m/s at an angle ?=44o (counterclockwise

Question

A ball "a" is shot with a speed u=1.4 m/s at an angle ?=44o (counterclockwise from east) towards a ball "b". Ball "a" hits ball "b" with the axis of impact lying along the x axis. After the collision, "b" is moving in the positive x direction and the "a" is moving in the positive y direction. (both balls have equal mass.)
(a) What is the velocity of the center of mass of the system of two balls before the collision?
( i + j) m/s

(b) What is the speed vb of the ball "b" after the collision?
m/s
(c) What is the speed va of the ball "a" after the collision?
m/s
(d) What is the velocity of the center of mass of the system of two balls after the collision?
( i + j) m/s

(e) Is the collision elastic, inelastic, or partially elastic?
More information is needed.
Inelastic.
Partially elastic.
Elastic.

Explanation / Answer

The diagonal of a square with sides 1 is 1.4(14). If we break the velocity up into x and y components, the speeds will both be 1 m/s. Could also find y with y = 1.4 sin 45 and x = 1.4 cos 45 For any type collision, the momentum is conserved, so the center of mass keeps going at the same velocity. a) before, it is moving at 1 m/s in the x direction and 1 m/s in the y direction. i is the unit vector in the x direction. j is the unit vector in the y direction 1(i+j) m/s b) after the collision "a" has zero side-to-side momentum so "b" must have the same side-to-side momentum that "a" had before the collision so vb = 1 m/s c) same logic as above, just flipped around va = 1 m/s d) CoM velocity is conserved (or just add up the 1m/s in each direction) v (CoM) = 1 (i+j) m/s e) 1/2 m (1m/s)^2 + 1/2 m (1m/s)^2 = 1/2 m (1.4 m/s)^2 so it is elastic. If it weren't totally elastic, there would be a component of a's velocity in the x direction and a component of b's velocity in the y direction.

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