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August 31, 2014

August 31, 2014

Posted by **jason** on Monday, December 22, 2008 at 11:28am.

Calculate the maximum height of ball 1 from and ground and the rebound height of ball 2.

2. Relevant equations

3. The attempt at a solution

can you check my work i am stuck

I would calculate the kinetic energy of ball 2 just before and after it hits the ground, using

Just before impact, ball 2 (M) and ball 1 (m) have acquired kinetic energy

(1/2)mV^2 + (1/2)MV^2

= M g*0.74 - m g*0.74 Solve for V of both balls, and the kinetic energy of ball 2.

(1/2)(m + M)V^2 = 0.74 (M-m)g

V^2 = [(M-m)/(M+m)]g*1.48 = 12.38 m^2/s^2

V = 3.52 m/s

= After impact, ball 2 will have 70% of the pre-impact kinetic energy, and its velocity will be V' = sqrt(0.7)V= 0.8367V = 2.94 m/s

Right after ball 2's impact, Ball 1 will continue to rise for a while because it suffered no impact and maintained its velocity V when ball 2 hits the ground. There will be no tension in the string while ball 2 and ba1l 1 both rise.

Ball 1 rises a distance H1 given by

M g H1 = (1/2) M V'^2 = (1/2)M(0.7)V^2

H1 = (0.7)V^2/(2g) = 0.35 V^2/g

When ball 1 hits the ground, ball 2 has already risen from 0.74 to 0.74 + 0.88 = 1.62 m above the ground. It will rise an additional distance until its kinetic energy (1/2) mV^2(at impact)is converted to potential energy

i am getting the wrong anwser

(1/2)mV^2 + (1/2)MV^2

= M g*0.74 - m g*0.74 Solve for V of both balls, and the kinetic energy of ball 2.

Shouldnt it be Mg*0.88 instead of Mg*0.74 for ball 2

- physics -
**drwls**, Monday, December 22, 2008 at 11:43amYes, you caught an error that I made. Good work. Ball 2 falls 0.88 m before it hits the ground, not 0.74 m. That affects the computed velocities V and V'. See if you can get the correct by answer making those changes

- lxbw jarcis -
**lxbw jarcis**, Sunday, February 1, 2009 at 7:37pmzpoifut gyched dribzgc imhnsgaw mczke rpaqw hjdsyr

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