A bullet of mass 0.117 kg traveling horizontally at a speed of 200 m/s embeds itself in a block of mass 2 kg that is sitting at rest on a nearly frictionless surface.
What is the speed of the block after the bullet embeds itself in the block?
v = m/s
2007-09-08 17:10:50 · 3 answers · asked by Elvis P 1 in Science & Mathematics ➔ Physics
Momentum is conserved
Initial Momentum =Final Momentum
m1*v1=m2*v2
0.117*200=(2+0.117)*v {as bullet embeds add both masses}
v=11.053 m/s
2007-09-08 17:17:21 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If the block of mass 2 kg is sitting on a frictionless surface, then the speed of the block will equal 200 m/s aafter the bullet strikes it. This is an impossible situation by the way!
2007-09-09 00:18:35 · answer #2 · answered by Timothy K 1 · 0⤊ 1⤋
mv=MV
It's just plug 'n chug.
The bullet has a momentum of .117*200 = 23.4 m-kg/s
After the (perfectly inelastic) collision, the system has the same momentum so
2.117*V = 23.4 => v = 11.05 m/s
HTH
Doug
2007-09-09 00:18:52 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋