Physics: Conservation of Momentum?

Question

1. A gun is fired vertically into a 2.00 kg block of wood at rest directly above it. If the bullet has a mass of 0.021 kg and a speed of 250 m/s, how high will the block rise into the air after the bullet becomes embedded in it?
2. A 1700 kg car moving south at 14.5 m/s collides with a 2500 kg car moving north. The cars stick together and move as a unit after the collision at a velocity of 5.46 m/s north. Find velocity of the 2500 kg car before the collision.
3. A moving object has a kinetic energy of 200 J and a momentum with a magnitude of 33 kg*m/s. Determine the mass and speed of the object.
4. A 68.3 kg student stands in the middle of a frozen pond having a radius of 5 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty he throws his 2.8 kg physics book horizontally toward the north shore at a speed of 6 m/s. How long does it take him to reach the south shore in seconds?

Answer 1

1. use mvi1+mvi2=vf(m+m)
2(0)+.021(250)=vf(2+.021) vf turns out to be 2.598.
then use v^2=v^2+2as, to get 0=2.598^2+2(-9.8)s, s turns out to be .344m
2. use same formula to get (1700)(-14.5)+(2500)vi=5.46(1700+2500). (south being negative)
vi turns out to be 19.03m/s
3. use both kinetic energy formula and the momentum formula. 200=.5mv^2 and mv=33. use substitution to get v to equal 12.12m/s. then put that back in the momentum formula to get mass to be 2.72kg
4. momentum is conserved, so 2.8(6)=68.3v. v turns out to be .246m/s. then use s=.5at^2+vit. acceleration is 0, so 5=.246t. turns out to be 20.33seconds

Answer 2

♠ bullet’s kinetic E=0.5*m1*v^2 being transformed into potential energy W= (m1+m2)gh, where m1=0.021kg, v=250m/s, m2=2kg, g=9.8m/s^2, h to be found;
energy conservation law: E=W;
thence h=0.5*0.021*250^2/(2.021*9.8) =33.13m; !???!
♣ momentum conservation law: m1*v1+m2*v2=(m1+m2)*v3, where m1=1700kg, v1=-14.5m/s (I take moving northward positive), m2=2500kg, v2 to be found, v3=+5.46m/s;
thence v2= {(m1+m2)*v3 –m1*v1}/m2 =
= ((1700 +2500)*5.46 +1700*14.5)/ 2500 =19m/s; OK!
♦ momentum 33=m*v; energy 200=0.5*m*v^2; 200/33=0.5*v, hence v=12.12m/s m=33/v =2.7225kg; OK!!
♥ momentum conservation law: 0=m1*v1+m2*v2, where m1=2.8kg, v1=-6m/s, m2=68.3kg; thence v2=-v1*(m1/m2); t =R/v2 =(5/6)*(68.3/2.8) =20.33s; OK!!