A 1.5 kg 8-Ball is traveling at 2.3 m/s to the right when it strikes and bounces off a 5.2 kg bowling ball which is at rest. After the collision, the 8-Ball moves to the left with a velocity of 1.1 m/s. What is the final velocity of the bowling ball? m/s (nearest hundredth)

-0.34m/s

Explanation

According to the law of conservation of momentum

m1u1 +m2 u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v s the common velocity

Given

m1 = 1.5kg

m2 = 5.2kg

u1 = 2.3m/s

u2 = -1.1m/s (left direction)

Required

Final velocity v of the ball

Substitute into the formula

1.5(2.3)-5.2(1.1) = (1.5+5.2)v

3.46-5.72 = 6.7v

-2.26 = 6.7v

v = -2.26/6.7

v = -0.34m/s

Hence the final velocity of the bowling ball is 0.34m/s towards the left

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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