A 200 g steel ball and a 400 g steel ball each hang from 2.0- m-long strings. At

Question

A 200 g steel ball and a 400 g steel ball each hang from 2.0- m-long strings. At rest, the balls hang side by side, barely touching. The 200 g ball is pulled to the left until the angle between its string and vertical is 39 degree. The 400 g ball is pulled to a 39 degree angle on the right. The balls are released so as to collide at the very bottom of their swings. To what angle does 200 g ball rebound? Express your answers using two significant figures. theta_200g ball = degree To what angle does 400 g ball rebound? Express your answers using two significant figures. theta_ 400g ball = degree

Explanation / Answer

we will use conservation of momentum,

height to which both the ball are raised= L ( 1- cos 39) = 2 ( 1-0.777) =0.4457 m apprrx

mgh = 1/2 mv^2

v= sqroot ( 2x 9,8 x 0.4457) = 2.95 m/s apprx, v^2 = 8.73

1/2 m1 v^2 + 1/2 m2v^2 = 1/2 m1 V1^2 + 1/2 m2 v2^2 ( conserving energy)

8.73 ( 600) = 200 V1^2 + 400 V2^2

8.73 (3) = V1^2 + 2V2^2-

26.19 = V1^2 +2V2^2---------eq(1)

V1= ( 200-400) (2.95)- 2(400)2.95/ 600 =-4.916 m/s

mg L ( 1-cos theta) = 1/2 mv^2

9.8 (2) (1- costheta) = 12.08

theta = 67.48 degree apprx

V2= ( 400-200 ) (-2.95) + 2 (200) ( 2.95)/ 600 = 0.98 m/s apprx

mg L ( 1-cos theta) = 1/2 mv^2

9.8 ( 2) ( 1-cos theta) = 0.5 (0,98)^2

thet a= 12.708 degree apprx

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