A 3.8 kg block of wood initially at rest is struck by a bullet moving at 494 m/s. The bullet exits the block of wood with a speed of 293 m/s.
a) If the block of wood moves in the same direction as the bullet with a speed of 1.9 x 10-2 m/s, what is the mass of the bullet?
b) How much kinetic energy was lost?
a) If the block of wood moves in the same direction as the bullet with a speed of 1.9 x 10-2 m/s, what is the mass of the bullet?
b) How much kinetic energy was lost?
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a) Initial Momentum of bullet =494m
Initial momentum of block = 0
Final momentum of bullet = 293m
Final momentum of block = 3.8 x 1.9x10^-2
Using conservation of momentum:
Total initial momentum = total final momentum
494m + 0 = 293m + .8 x 1.9x10^-2
It's just a case of solving this for 'm', the mass of the bullet, which is best if you do for yourself!
b) Work out the initial KEs of the block and of the bullet (use KE = mv^2/2 for each and add)
Work out the final KEs of the block and bullet and add them,
KE lost is just the difference between the total initial KE and the total final KE.
Initial momentum of block = 0
Final momentum of bullet = 293m
Final momentum of block = 3.8 x 1.9x10^-2
Using conservation of momentum:
Total initial momentum = total final momentum
494m + 0 = 293m + .8 x 1.9x10^-2
It's just a case of solving this for 'm', the mass of the bullet, which is best if you do for yourself!
b) Work out the initial KEs of the block and of the bullet (use KE = mv^2/2 for each and add)
Work out the final KEs of the block and bullet and add them,
KE lost is just the difference between the total initial KE and the total final KE.
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Momentum lost by bullet = P1 = m(494 - 293) = 201*m
201*m = momentum gained by block = M*Vb = 3.8*.019 = .0722 kg∙m/s
a) m = .3592 g
b) KElost = KE2 - KE1 = ½m*293² + ½*3.8*.019² - ½m*494² = -28410 J
201*m = momentum gained by block = M*Vb = 3.8*.019 = .0722 kg∙m/s
a) m = .3592 g
b) KElost = KE2 - KE1 = ½m*293² + ½*3.8*.019² - ½m*494² = -28410 J