1)A 48 g ball is fired horizontally with initial speed v 0toward a 100 g ball th

Question

1)A 48 g ball is fired horizontally with initial speed v0toward a 100 g ball that is hanging motionless from a 1.1 m -long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle max = 51 .

What was v0?Express your answer using two significant figures

2)A 59 g ice cube can slide without friction up and down a 35 slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 12 cm . The spring constant is 23 N/m .

When the ice cube is released, what distance will it travel up the slope before reversing direction? (express in 2 sig figs)

3)A 65 kg skateboarder wants to just make it to the upper edge of a "half-pipe" with a radius of 3.0 m ,

First, treat the skateboarder and board as a point particle, with the entire mass nearly in contact with the half-pipe. What speed v0 does he need at the bottom if he is to coast all the way up?

More realistically, the mass of the skateboarder in a deep crouch might be thought of as concentrated 0.60 m from the half-pipe. Assuming he remains in that position all the way up, what v0 is needed to reach the upper edge ?

Explanation / Answer

In accordance with law of conservation of energy Kinetic energy with which ball hit the hanging ball both are brought up to certain height. Therefore, Kinetic energy of the moving ball will equal to potenatil energy of the both the balls

Expression for Kinetic energy

K.E = 1/2 m V02

Potenatil energy of the both the ball ,when they at vertical height

P.E =(M+ m)gh

Here ,h is the vertical height reached by the both the

balls : h = (L - Lcos)

Apply the law of conservation of enegy

1/2 m V02   = (M+ m)g(L - Lcos)

1/2 (0.048 kg) V02 = (0.10 kg+ 0.048 kg) (9.8) (1.1 m- 1.1 m * cos51o)

Solve the equation for  V0

V0 = 4.964 m/s

Round off the result to two significant digits

V0 = 5.0 m/s  

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