Question

1. A collision in which two bodies remain joined together after the collision is referred to as.

2. A 0.5 kilogram cart moving at 0.360 m/s has how much momentum

3. If the 0.5kg cart moving at 0.36 m/s was to hit a barrier and bounce back with a velocity of - 0.240 m/s, what is its change in momentum

4.How fast must a 0.25 kg cart be traveling to have a momentum of 0.450 kgm/s?

5. A 0.230 kg baseball is thrown with a speed of 41 m/s. If the ball comes to rest in the catcher’s mitt in 0.085 seconds, how much force does the ball apply on the catcher’s mitt? (hint: use the impulse-momentum theorem )

6. Imagine you are ice skating with your friend. Both of you are at rest, when you shove him/her away from you. You have a mass of 65 kg and he/she has a mass of 55kg. When you shove off, you move away with a velocity of 2.0 m/s. With what velocity does your friend move away from you?

7. If a 0.25 kg cart moving to the right with a velocity of +0.31 m/s collides inelastically with a 0.5 kg cart traveling to the left with a velocity of - 0.22 m/s, what is the total momentum of the system before the collision?

8. What is the resulting velocity of the above two-car system (stuck together)?

Answer 1

1)

A collision in which two bodies remain joined together after the collision is called as inelastic collision.

Because in inelastic collision the total momentum of  two bodies remains the same.

=========================================================

2)

m = 0.5 kg

v = 0.36 m/s

momemtum(P) = mass*velocity

= 0.5*0.36

= 0.18 kgm/s

==============================================================

3)

momentum(P1) = 0.18

m = 0.5 kg

v = -0.240 m/s

momentum(P2) = m*v

=0.5 * -0.240

= - 0.12 kgm / s

Change in momentum = P2 - P1

= - 0.12 - 0.18

= - 0.3 kgm / s

================================================================

4)

momentum = 0.450 kgm/s

mass = 0.25 kg

velocity = momentum / mass

= 0.450 / 0.25

=1.8 m/s

=============================================================

5)

mass = 0.23 kg

velocity = 41 m/s

change in time (delta t) = 0.085 sec

Impulse = force * change in time

change in momentum = m*v

= 0.23*41

=9.43 kgm/s

Force = delta P / delta t

= 9.43 / 0.085

= 110.941 N

==========================================================

6)

m1 = 65 kg

m2 = 55 kg

V1 = 2 m/s

m1 * v1 = m2 * v2

V2 = m1 * v1 / m2

= 65 * 2 / 55

= 2.364 m/s

==============================================================

7)

m1 = 0.25 kg

v1 = 0.31 m/s

m2 = 0.5 kg

v2 = - 0.22 m/s

total momentum of the system before collision = m1*v1 + m2*v2

= (0.25 ) (0.31) +( 0.5) (-0.22)

= 0.0775 - 0.11

= - 0.0325 kgm /s

====================================================================

8)

In above problem

total momentum( m1*v1 + m2*v2) = - 0.0325 kgm/s

m1*v1 + m2*v2 = (m1 + m2)* V

- 0.0325 =( 0.25 + 0.5)* V

V = (- 0.0325) / (0.75)

= - 0.0433 m/s