The initial momentum of the steel ball at the bottom of theramp is determined by

Question

The initial momentum of the steel ball at the bottom of theramp is determined by letting a ball roll from a height, h abovethe ramp. As it rolls down the ramp, its speed increases. The speedat the bottom of the ramp can be determined by applying theprinciple of conservation of energy since there is very littleenergy lost due to rolling friction. At height, h, the totalmechanical energy is the potential energy, mgh and at the bottom ofthe ramp, all the potential energy is converted into thetranslation (½mv2) and rotational(½I2) kinetic energies of theball.
the horizontal distance the ball traveled after it leaves theramp is x0 = vbt. For any object which has noinitial vertical velocity, the time it takes to fall from theheight H is always the same, independent of the value of theinitial horizontal velocity. This implies that thehorizontal distance from the point where the ball leaves the rampis proportional to the initial horizontal velocity, and is thusalso proportional to the momentum.
For this exercise, the mass of the ball is 25.3 g, the heighth = 11.5 cm and H = 41.6 cm.
1 What is the horizontal velocity of the ball at thebottom of the ramp? (in m/s)

2 What is the momentum of the spherejust before it leaves the bottom of the ramp? (inkg*m/s)

3 How long does it take for the ball to fallfrom the bottom of the ramp to the floor? (s)

4 What is the horizontal distance,x0 traveled by the ball? (cm)

Explanation / Answer

(1) . mass m = 25.3 g                   = 0.0253 kg height h = 11.5 cm = 0.115 m          H = 41.6 cm =0.416 m Let Horizontal velocity of the ball at the bottom of theramp be v Then from law of conservation of energy , ( 1/ 2) m v ^ 2 * [1+ ( k/ r ) ^ 2 ] = m g h for steel ball ( k / r ) ^ 2 = 2/ 5 substitue above we get , ( 7/ 10 )v ^ 2 = gh from this v = [ gh / 0.7 )                 = 1.2688 m / s (2). the momentum of the sphere just before it leaves thebottom of the ramp p = m v                                                                                                                       = 0.0321 kg m / s (3) . Required time t = [ 2H / g ]                                = 0.291 s                                =

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