help!! physic questions....!plz help!!?

Question

Plz help me thx..

1) what is the impulse due to a force rthat causes the speed of a 46 g golf ball to change from 0 to 60.0 m/s in 0.50ms?

2) A 0.250 kg ball of Plasticine moving 5.0 m/s overtakes and collides with a 0.3000 kg ball of Plasticine, traveling in the same directiion at 2.0 m/s. The 2 balls of Plasticine stick together pm collison. What is their speed after the collision.

3) A 4.2 kg rifle shoots a 0.05 kg bullet at a speed of 3.00 * 10^2 m/s. At what speed does the rife recoil?

4) A railroad car of mass 12,000kg is traveling at a speed of 6.0 m/s when it collies with an identical car at rest. Thte two cars look together. What is their common speed after the collision?

5) What impuse is needed to change the speed of a 10 kg object from 1.6 m/s to 25.5 m/s in a time of 5s. How much force is needed?

It would be awesome to show the steps...
Once again, ty so much.........

Answer 1

1.m(mass of ball)=46 g.
=(46/1000) kg.
u(initial velocity)=0
v(final velocity)=60m/s
Impulse=Change of momentum
=Final momentum-Initial momentum
=mv-mu
=m(v-u)
=(46/1000)X(60-0)
=2.76 Ns.

2.m1(mass of first plasticine)=0.250 kg
m2(mass of second plasticine)=0.3 kg
v1(velocity of first plasticine)=5 m/s
v2(velocity of second plasticine)=2 m/s
After collision let the combined mass be M=m1+m2
=0.25+0.3
=0.55 kg
Applying law of conservation of momentum,
Initial momentum=Final momentum.
m1v1+m2v2=MV (where V is the final velocity of the system)
(0.25X5)+(0.3X2)=0.55V
1.25+0.6=0.55V
V=(1.25+0.6)/0.55
V=3.36 m/s.

3.m(mass of bullet)=0.05 kg
M(mass of rifle)=4.2 kg
v(velocity of bullet)=300m/s
Let V be the recoil speed of rifle.
Aplying law of conservation of mmomentum,
Initial momentum=Final momentum.
mv+MV=0
(0.05X300)+(4.2V)=0
15+4.2V=0
V=-(15/4.2)
V=-3.57m/s.
The negative sign indicates that the direction of recoil is opposite to the direction of motion of the bullet.

4.m1(mass of the first car)=12000 kg
m2(mass of the second car)=12000 kg(because both the cars are identical)
v1(velocity of first car)=6 m/s
v2(velocity of second car)=0(because it is at rest)
Let M be the combined mass after collision.
M=m1+m2
=12000+12000
=24000 kg
Applying law of conservation of momentum,
Initial momentum=Final momentum.
m1v1+m2v2=MV
(12000X6)+(12000X0)=24000V
72000+0=24000v
V=(72000/24000)
V=3 m/s.

5.m(mass of the object)=10 kg
u(initial velocity)=1.6 m/s.
v(final velocity)=25.5 m/s.
t(time taken)=5 s.
a(acceleration)=(v-u)/t
=(25.5-1.6)/5
=4.78 m/s^2.
F(force)=mXa
=10X4.78
=47.8 N.
I(impulse)=FXt
=4.78X5
=23.9 Ns.

Answer 2

1)
Impulse is given by F*∆t and is equal to momentum change ∆(m*v). Since the mass is constant, the momentum change is m*∆v. You are given ∆v = 60 m/s, and ∆t = 0.5*10^-3 s. m*∆v = 0.046 kg * 60 m/s = 2.76 kg*m/s. You are not asked for the force, but you can get that by dividing the impulse by ∆t: F = 5.52*10^3 kg*m/s^2 = 5.52*10^3 N.

2) Conservation of momentum: equate the system momentum before and after the collision. Momentum before: 0.250*5.0 + 0.300*2.0. Momentum after: (0.25 + 0.300)*v, where v is the final velocity. Equate these and solve for v:

v = (0.250*5.0 + 0.300*2.0) / (0.300 + 0.25) = 3.36 m/s

3)
Another conservation of momentum problem. Before the firing, the system momentum is zero, therefore the momentum of the bullet (0.05*3*10^2) must equal the momentum of the rifle (they are in opposite directions, so they add to zero) which is 4.2*v. Again, equate these and solve for v:

v = (0.05*3*10^2)/4.2 = 3.57 m/s

4)
Do this the same way as 2)

5)
Impulse = m*∆v = 10*(25.5 - 1.6) = 239 kg*m/s

The force is impulse/∆t = 239/5 = 47.8 N

Answer 3

1. The impulse is simply 0.046 x 60 kg-m/sec. The time is irrelevant for this, although highly relevant if one were computing the force.
2. In this problem, momentum is conserved (although energy is not). Add the momenta of the two objects, and divide the sum by the combined mass.
3. This is also a conservation of momentum problem; it is a simple proportion.
4. 3 meters per second. Again, a conservation of momentum problem, particularly simple in this case.
5. The impulse is 10 x (25.5 - 1.6) kg-m/sec. Divide by 5 to get the force in newtons.

Answer 4

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