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Hi.

I have a bit of a dillema and a few problems regarding a simple excersize.

The situation is as folows. We have a 6,00 g bullet that has a K.E. of 3.12 J. It strikes a 122 g cart that is on railway tracks. The bullet hits the cart parallel to the tracks. Now we need to find out 1) Speed of bullet before impact 2) Speed of cart after impact 3) With what impulse did the bullet act on the cart 4) How much K.E of the bullet has converted in the internal energy of the bullet 5) For how much deg. K would the cart and bullet be heated if they are from lead with Specific heat = 130 J /KG K and the heat would not be transimitted to the environment 7) For how much would the cart and bullet be heated if the bullet would hit the cart perpendicular to the railway track.

I have difficulty with 6 and 7 and also a bit with 3.

1) Plain enough. K.E = 1/2 m * v^2 with some basic algebra we find that v = 32,2 m/s

2) Momentum before = Momentum after

m(bullet before)*v(bullet before) + 0(cart has no momentum before) = m(bullet)*v + m(cart)*v

6*32,2 = 6v + 122v with some algebra we find that v = 1.5 m/s[/quote]

Yes, looks good. Since you used grams through out, you don't need to convert to kg.

I have a bit of a dillema and a few problems regarding a simple excersize.

The situation is as folows. We have a 6,00 g bullet that has a K.E. of 3.12 J. It strikes a 122 g cart that is on railway tracks. The bullet hits the cart parallel to the tracks. Now we need to find out 1) Speed of bullet before impact 2) Speed of cart after impact 3) With what impulse did the bullet act on the cart 4) How much K.E of the bullet has converted in the internal energy of the bullet 5) For how much deg. K would the cart and bullet be heated if they are from lead with Specific heat = 130 J /KG K and the heat would not be transimitted to the environment 7) For how much would the cart and bullet be heated if the bullet would hit the cart perpendicular to the railway track.

I have difficulty with 6 and 7 and also a bit with 3.

1) Plain enough. K.E = 1/2 m * v^2 with some basic algebra we find that v = 32,2 m/s

2) Momentum before = Momentum after

m(bullet before)*v(bullet before) + 0(cart has no momentum before) = m(bullet)*v + m(cart)*v

6*32,2 = 6v + 122v with some algebra we find that v = 1.5 m/s[/quote]

Yes, looks good. Since you used grams through out, you don't need to convert to kg.

Yes.3) Now the impulse of the bullet on the cart.

Since the momentum of the cart is changed from 0 to m(cart)* 1,5 m/s = 0.183 KG m /s. The change of momentum of the cart is then 0.183 Kg m/s. Then the impulse of the bullet on the cart is equal to the momentum change of the cart = 0.183 Kg m/s. Would my reasoning be correct?

No. Both cart and bullet are moving at 1.5 m/s. The mass is 6+ 122= 128 g= 0.128 kg. The total kinetic energy after the bullet hits the cart is 1/2*(0.128 kg)(1.5 m/s)4) The K.E(before) = K.E(after) + Internal E.(after)

3.12 J = 1/2 * (6g) * (1.5 m/s)^2 + x would this reasoning be correct? We find that x = 3.11. Total I.E = 3.11 J. Soo we have that 3.11 J of E has converted in the internal energy. Correct?

^{2}= 0.62 J. That means that 3.12-0.62= 3.5 J was converted to heat.The correct E is 3.5 J.5) Now with this I am puzzled. I know that E= m * cp(specifci heat) * delta T. But what value should I use for E. I recon. That 3.11 J of Internal goes to heat. Soo we have delta T = 3.11 J / (6g + 122 g) * 130 J/KgK. delta T = 0.18 K. Correct?

Yes.6) For this all of the K.E would go to I.E since neither the bullet nor the cart are moving after the impact. I recon. delta T = 3.12 J / (6g+ 122g) * 130 J/KgK. Delta T = 0.19 K.

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