# Need help with a couple of physics questions



## Flame Sixtyone (Aug 27, 2009)

I need help with 2 questions (maybe this isnt the best place to ask) dealing with Work, Energy, and Momentum 

1. An 8 kg puck floating on an air table is traveling east at 15 cm/s. Coming the other way at 25 cm/s is a 2 kg puck on which is affixed a wad of bubble gum. The slam head-on into each other and stick together. Find their velocity after the impact. How much kinetic energy is lost?

2. A soft clay block is suspended so as to form a ballistic pendulum. A bullet is fired point-blank into the block, imbedding itself therein and raising the latter to a height h. Write an expression for the muzzle speed of the bullet in terms of g, h, and the masses. Remember that although friction on the pendulum is negligible, clay-bullet friction losses are not, and mechanical energy is not conserved in the collision

I looked at the back of the textbook, and the answer to the first question is 0.07 m/s and 0.13 J. 
the answer to the second question is 
Vb = ((mb + mc)/mb)*sqrt(2gh)
, where vb is the velocity of the bullet, mb is the mass of the bullet, and mc is the mass of the clay block, g is the acceleration due to gravity, and h is the height at which the clay block was raised when the bullet hit it

but it doesnt give the method on how to solve them. help!


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## Citrine (Oct 19, 2010)

well for the first one, we know that momentum is conserved, whether it is an inelastic or elastic collision. Therefore....

initial momentum=final momentum
*p*_initial=_*p*_final_
*m*_1_*v*_1_* + m*_2_*v*_2_* = (m*_1_* + m*_2_*)**v*_final_

plug in the numbers and solve for the final velocity. Remeber to convert the velocities from cm to m.

The kenetic energy that is lost is the change of kenetic energy. You'll need know that kinetic energy = (.5)(mass)(velocity)squared. the change of the kinetic energy is final KE - initial KE. For the final KE use the velocity you just found. For the inital KE, add the kinetic energy of both pucks before the collision. Hope this helped!


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## rickey (Jun 22, 2010)

For second question,

First apply conservation of momentum to the whole system. You should have an equation like this for the final velocity,

vf = m1vb/(m1+m2)

When work and energy is concerned, you need to take into account that there exists a gravitational potential energy and kinetic energy.

INITIALLY - both the bullet and clay's height dont change and as a whole system, the kinetic energy is zero (even though the bullet has an initial speed).

FINAL SITUATION - both the bullet and clay's height has changed (to the same height). So the final potential energy for the system is: (mb+mc)gh. They're both moving at the same final speed so the final kinetic energy is: 1/2(mb+mc)vf^2. Remember to set these equal to each other (work-energy theorem). But the question asks for the initial bullet speed. Luckily, we have vf from our momentum equation which has vb. Just plug this equation into the work-energy equation and with a little algebra, you get the answer the book gives you. TADA!!


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## Flame Sixtyone (Aug 27, 2009)

Thank you both of you! that really helps a lot. I especially had no idea how to do the second problem
In the first problem I found out that *v*_2_ should be negative since its a vector and its in the opposite direction of positive *v*_1_. I tried it and it worked


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