Getting two Answers : calculating velocity of ball

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Mr. Gale's experiment to measure the velocity of a ball using a box sliding on a surface yielded two different results from two methods of calculation. The first method, based on energy conservation, calculated the ball's velocity as 6.0 m/s, while the second method, using force and momentum, resulted in a velocity of 14.64 m/s. The discrepancy arises from the assumption that all kinetic energy is lost during the sliding phase, which is not accurate as some energy is released during the impact. It was clarified that method two is correct and can be used to determine the energy lost in the collision. This discussion highlights the complexities of energy conservation and momentum in collision scenarios.
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Homework Statement



Mr. Gale decides to build his own velocity measuring machine.
His plan is to build a box (mass 2.50 kg) that catches the ball (mass 0.500 kg) being thrown. It then slides across the desk but eventually comes to a stop due to friction (μ = 0.800) between the box and the surface. He thinks that knowing how far the box slides will allow him to calculate the velocity of the ball just before it hits the box. To test this out he throws the ball and the box slides exactly 38 centimeters. Calculate how fast the ball must have been going just before it hit the box.

The Attempt at a Solution



There were two different ways to approach it. But this gave different answers. One of these is wrong.

ATTEMPT 1 : Using Energy
KE of the ball before just before it hits the box = Work done by friction on the box and ball combined

0.5(0.5kg)v^2 = (0.800)(2.50 + 0.5)(9.8)(0.38)
v = 6.0 m/s

ATTEMPT 2 : Using Force and Momentum
friction is net force, calculate velocity, use momentum to calculate initial velocity in ball

f = (2.50 + 0.5)(9.8)(0.800) = 23.52 = ma
therefore a = (23.52)/(2.50 + 0.5) = 7.84 m/s^2
using v^2 = 2ad
v = sqrt((2)(7.84)(.38)) = 2.44 m/s

therefore pi = pf
(0.5)(v) = (2.44)(2.50 + 0.5)
v = 14.64 m/s

why am I getting different values? There is something I am forgetting here.
Thanks for any help in advance
 
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Is work conserved in the process of catching the ball?
 
i think the assumption i am making is that

KE + Wnc = 0
 
fishspawned said:
i think the assumption i am making is that

KE + Wnc = 0
There are two stages here. In the first stage, the ball is caught by the box, in the second the box slides. You have assumed all initial KE is lost in the sliding stage. Is that a safe assumption?
 
From what you are saying, I am taking it that the impact of the ball on the box would release some energy before it even begins to slide? Could this be calculated? I assume not without further information. However, now this leads me to even more confusion. Are you suggesting that method 2 is correct?
 
fishspawned said:
From what you are saying, I am taking it that the impact of the ball on the box would release some energy before it even begins to slide? Could this be calculated? I assume not without further information. However, now this leads me to even more confusion. Are you suggesting that method 2 is correct?
Method 2 is correct. You can use it to find how much energy was lost in the impact.
Note that for no energy to be lost in the collision it would have to be completely elastic. Newton's "experimental" law tells you that the relative velocity would be negated. That is, the ball and box would move apart just after the collision as fast they approached each other just before the collision.
 
Fantastic. The different velocities is giving me a way to calculate the actual energy loss from the impact. That's awesome. This turned out better than i thought. Many thanks for the help.
 

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