Physics question p4?

67) 

Energy given to the block is stored a spring potential.E = PS = ½kx² = ½(900)(0.05²) = 1.125 JThis energy was kinetic energy of the block meaning the initial velocity of the block 1.125 = ½(1.00)v²v = 1.50 m/sMeaning the block had a momentum of mv = 1.00(1.50) = 1.50 kg•m/sThis momentum was given up by the bullet, so subtracting it from the original bullet momentum allows us to determine the bullet exit velocity.0.005(400) - 1.50 = 0.005vv = 100 m/s71) We can use F = ma on each mass. For the one on the slopeT₁ - 15.0(9.81sin37.0) = 15.0(2.00)T₁ = 118.6 N

T₂ - 20.0(9.81) = 20.0(-2.00)

T₂ = 156.2 N

Now we can apply the angular equivalent 

τ = Iα

(T₂ - T₁)R = I(a/R)

I = (T₂ - T₁)R²/a = (156.2 - 118.6)(0.250²) / 2.00 = 1.175 kg•m²

67.  The bullet gives the block two things a) energy and b) momentum.

The energy given to the block causes it to continue to move until the spring stops it.  Ek = 1/2 k x^2 = 1/2 * 900 * 0.05^2 = 1.125 J

Now you need to calculate the speed of this block to find how much momentum it was given.  This momentum comes from the bullet.

1/2 * 1 * v^2 = 1.125  -> v = sqrt( 1*1.25)  = 1.118 m/s and its momentum was mv = 1.118 kgm/s

The bullet loses this momentum.

So now find the new speed of the bullet.  From that find a) its intial energy, b) the new energy after it loses its momentum  and of course you already know the energy of movement of the block so the "missing energy" is gone into internal energy in the block and the bullet.

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