Physics: Inelastic Collisions in One Dimension?

Question

49. A 3.50 g bullet is fired horizontally at two blocks at rest on a frictionless table. The bullet passes through block 1 (mass 1.20 kg) and embeds itself in block 2 (mass 1.80 kg). The blocks end up with speeds v1 = 0.630 m/s and v2 = 1.40 m/s. Neglecting the material removed form block 1 by the bullet,
Find the speed of the bullet as it:
(A) leaves and
(B) enters block 1.

The answers are:

(A) 721 m/s
(B) 937 m/s

But I don’t know how to get these answers, please help me!

Answer 1

Okay, figured it out. It took me a while because I haven't done any of this Physics in a while, but it is quite simple. You want to start with the second block, and work your way back to the beginning, when the bullet was fired.

This is a problem of conservation of momentum.
P1 = P2
before the bullet collides with the second block:
P1 = (M_bullet)*(V_bullet) + (M_block2)*(V_block2)
V_block2 is zero since it is at rest, therfore
P1 = M_bullet * V_bullet ; V_bullet is unknown and same as what part A is asking

P2 = (M_bullet + M_block2)*(V_bulletblock2combination)

The only unknown is V_bullet, so set P1 = P2, and solve for V_bullet. Be sure to convert the mass of the bullet to kg, and then you should get the answer of V_bullet = 721.4 (m/s).

I didn't do part B, but it should require the same logic. This is my first time doing this, so I'm sorry if it was verbose, and I hope it makes sense, nonetheless.

if you have any more questions, feel free to ask me at [e-mail]

-Ryan

Answer 2

The key to solving this is to remember that momentum in a closed system is always conserved.

The initial momentum of the system is just the mass of the bullet times it's initial velocity. Neither of the blocks is moving initially, so they have momentum of zero.

So initial momentum = (0.0035kg)v

The final momentum of the system is due to the motion of the first block, and the motion of the combined bullet and second block

Final momentum = (1.20)(0.63) + (1.8035)(1.40) = 3.28 m-kg/sec

Since momentum is conserved, initial momentum must equal final momentum.

(0.0035)v = 3.28

B) Initial velocity of bullet = v = 3.28/0.0035 = 937 m/s

To find the answer to A, we ignore the first block and just do initial and final momentum for the bullet striking the second block.
Its exactly the same sort of calculation, so I'll leave it to you

Answer 3

for completely inelastic collision common velocity V = m1u1 + m2u2 / m1+m2 ie., 0 = m v1 + 2m v2 / m + 2m ie., v1 /v2 = - (1/2) = - 1 : 2

Answer 4

Conservation of momentum: VB and VA are the answers you're looking for........

mb*(VB -VA) = m1*V1

mb*(VA-V2) = (mb + m2)*V2

Solve these 2 eqns simultaneously for VA & VB

Answer 5

A)
M1V1 = (M1 + M2)V2
V1 = (M1 + M2)V2/M1
V1 = (3.50 + 1800)(1.40)/3.50
V1 = 721.4 m/sec

B)
M1V11 = M1V12 + M2V22
V11 = (M1V12 + M2V22)/M1
V11 = (3.5*721.4 + 1,200*0.630)/3.5
V11 = 937.4 m/sec