4. A 3.00 kg mud ball has a perfectly inelastic collision with a second mud ball

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i need an answer i don't have much time 

May 17th, 2015

KE is 1/2*m*v^2, 150 = 1/2*m*v^2 
Momentum is m*v = 30 
so that's two eq with two unknowns. if you express v as 30/m and plug into the KE eq 
150 = 1/2*m*(30/m)^2 
solve for m, and then use the momentum eq to solve for v 

by conservation of momentum m1*v1 = m2*v2 
so 3.0*v = m2*v/3 
rearrange to 3.0*v*3/v = m2 
cancel the v and multiply 
3.0*3 = 9.0 = m2 

wow, complicated. first you have to find the time that the block is falling. since the vertical component of acceleration is only due to gravity, 
distance = 1/2*a*t^2 
1.5m = 1/2*9.81*t^2 
t=.553 secs 

so the horizontal velocity of the block was the horizontal distance traveled divided by the time 
2.5m / .553s = 4.52m/s 

so now were finally at the conv of momentum problem and m1*v1 = m2*v2 
5.5g*v1 = (5.5+22.6) * 4.52 
solve for v1, the speed of the dart 

It appears you have to find the distance that the force is applied thru. since the velocity changes by 8m/s, the average velocity through the collision is (v1+v2)/2 = 14m/s. 
the distance traveled in 5.0x10^-4s is v*t 
The energy change was 1/2m1v1^2-1/2m2v2^2. 
the energy change = force * distance, 
so force = energy change / distance.

May 17th, 2015

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