1. A collision in which two bodies remain joined together after the collision is
ID: 1333266 • Letter: 1
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)?
Explanation / Answer
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.
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2)
m = 0.5 kg
v = 0.36 m/s
momemtum(P) = mass*velocity
= 0.5*0.36
= 0.18 kgm/s
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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
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4)
momentum = 0.450 kgm/s
mass = 0.25 kg
velocity = momentum / mass
= 0.450 / 0.25
=1.8 m/s
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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
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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
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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
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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
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