# Bullet through Block

1. Oct 26, 2004

### SnowOwl18

I'm having some difficulties with this problem, even though it seems easy.

-----A bullet with a mass of 3.20 g is fired horizontally at two blocks resting on a smooth and frictionless table top as shown in the Figure. The bullet passes through the first 1.05 kg block, and embeds itself in a second 1.00 kg block. Speeds v1 = 4.60 m/s and v2 = 4.60 m/s, are thereby imparted on the blocks. The mass removed from the first block by the bullet can be neglected. Find the speed of the bullet immediately after emerging from the first block. Find the intial speed of the bullet.------

The second part I assume will be easier to figure out once I get the first part. For the first part I thought I could use the equation m1vi1 + m2vi2 = m1vf1 + m2vf2....but I keep trying to do it and I get lost and get wrong answers. Could anyone be so kind as to point me in the right direction? Thanks )

2. Oct 26, 2004

### HallsofIvy

Staff Emeritus
Yes, conservation of momentum is the way to go. There is no conservation of energy since the bullet becomes imbedded in the second block.

The bullet has mass 0.0032 kg and speed v. The two blocks are motionless so the total momentum is 0.0032v.

Afterwards, The block with mass 1.05 kg has speed 4.6 m/s and so has momentum (1.05)(4.6)= 4.83 kg m/s. The second block, together with the bullet imbedded in it, has mass 1.0032 kg. and speed 4.6 m/s so momentum 1.0032(4.6)= 4.61472 kg m/s.
That is, afterwards, the total momentum is 4.82+ 4.61472= 9.444 kg m/s. Conservation of momentum gives 0.0032v= 9.444 kg m/s. Solve that for v.

3. Oct 26, 2004

### SnowOwl18

ah that makes sense...so that is the initial speed of the bullet. now i just need to figure out the speed after it emerges from the first block.

Ah ha...I got it. I just kept setting up my equation incorrectly. Lol. Thanks for your help. )

Last edited: Oct 26, 2004
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