Ball 1, with a mass of 120g and traveling at 15.0m/s , collides head on with bal

Question

Ball 1, with a mass of 120g and traveling at 15.0m/s , collides head on with ball 2, which has a mass of 320g and is initially at rest. (a) What are the final velocities of each ball if the collision is perfectly elast? (b)What are the final velocities of each ball if the collision is perfectly inelastic?

A 50.0g ice cube can slide without friction up and down a 35.0? slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 12.0cm . The spring constant is 22.0N/m .A 50.0g ice cube can slide without friction up and down a 35.0? slope. The ice cube is pressed against a spring at the bottom of the slope, compressing the spring 12.0cm . The spring constant is 22.0N/m . (c) When the ice cube is released, what distance will it travel up the slope before reversing direction?

Explanation / Answer

The pendulum will swing to the same height/angle if the frictional force is neglected. As the energy dissipation is not mentioned so it should reach 23 degrees.

The height of fall is h= L (1-cos 23) = 0.5 (1-cos 23 ) = 0.03974m

Using conservation of energy KE = PE so (1/2) m V^2 = m g h , derive V = sqrt [ 2 g h] = 0.883m/s

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Spring constant problem

weight = kx so mg = kx or k = m g / x = 0.1kg*9.8 / 0.02m = 49N/m

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How much energy can be stored in a spring with k = 450N/m if the maximum possible stretch is 18.0cm?

A spring is compressed 1.8cm. (e) How far must you compress a spring with twice the spring constant to store the same amount of energy?

Some confusion here but will do that

d) E = (1/2) kx^2 = (1/2)(450)(0.18)^2 = 7.29J

e) twice the spring constnat is 2*450 = 900 now using hte same equation 7.29 = (1/2)(900)x^2 so x = 0.127m or 12.7cm

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(f) What was the speed of the cart just before it hit the spring?

(1/2)mV62 = (1/2) kx^2 or V = sqrt [ k x^2 / m ] = sqrt [ (230)(0.65)^2 /11] = 2.97m/s or 3m/s

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What are the final velocities of each ball if the collision is perfectly elast?

Momentum & energy conservation

m1 Vi1 + 0 = m1 Vf1 + m2 V2f and (1/2) m1 Vi1^2 + 0 = (1/2)m1 Vf1^2 + (1/2)m2 V2f^2

solving we get Vf1=(m1-m2)Vi1 / (m1+m2) = ( 0.12-0.32)15 /(0.12+0.32) = - 6.82 m/s reverse direction (minus sign)

Vf2 = 2 m1 Vi1 /(m1+m2) = 8.2 m/s forward direction

What are the final velocities of each ball if the collision is perfectly inelastic?

  m1 Vi1 + 0 = (m1 + m2) Vf so Vf = 0.12 (15) / (0.12+0.32) = 4.1 m/s

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(i) When the ice cube is released, what distance will it travel up the slope before reversing direction?

Energy conservation

(1/2) mV^2 = (1/2) k x^2 so V= sqrt [ kx^2 / m ] = 2.52m/s is the initial speed at the bottom.

acceleration a = g sin theta = 9.8 sin 35 = 5.621m/s^2

now distance is Vf^2 - Vi^2 = 2 a X . Here Vf = 0 , reversing direction then X = Vi^2 / 2 a = 0.5636m or 56.4cm

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(j) If the collision is perfectly elastic, what is the speed of the 120g ball after the collision? (to left or right)? (k)If the collision is perfectly elastic, what is the speed of the 400g ball after the collision? (to the let or right?)

Momentum & energy conservation

m1 Vi1 + m2Vi2 = m1 Vf1 + m2 V2f and (1/2) m1 Vi1^2 + (1/2)m2Vi2 = (1/2)m1 Vf1^2 + (1/2)m2 V2f^2

solving we get Vf1={[(m1-m2)Vi1}+[2m2 Vi2]} / (m1+m2) ={[( 0.12-0.4)4.4+[2*0.4*1.1] /(0.12+0.4) = - 0.677 m/s left direction (minus sign)

Vf2 ={[2 m1 Vi1]+[(m2-m2)Vi2]} /(m1+m2) = 2.623 m/s right direction

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