Collisions

Review

Momentum is a quantity of motion.p = mv

A change in momentum is called impulse.Impulse = p = mv

Impulses are caused by forces applied over time.p = Ft

Collisions

Collision – an isolated event in which 2 or more bodies exert strong forces over short periods of time against each other. Momentum is conserved in all collisions

(assuming no outside force). “sticky” collision – objects stick together

after colliding and move as one. “non-sticky” collision – objects bounce apart

after colliding.

Collisions

A train car with a mass of 3.00x104 kg traveling north at 1.5 m/s collides and couples with a 3.20x104-kg train car going south at 0.80 m/s. What is the velocity of the coupled

cars after the collision?

Collisions

Sketching the situation can help.

North

1.5 m/s1.5 m/s -0.80 m/s-0.80 m/s

mass = 3.20x104 kgmass = 3.00x104 kg

p1 = m1v1 p2 = m2v2

p1 = (3.00x104 kg)(1.5 m/s) p2 = (3.20x104 kg)(-0.80 m/s)

p1 = 4.5x104 kg*m/s p2 = -2.6x104 kg*m/s

ptotal = (4.5x104 kg*m/s) + (-2.6x104 kg*m/s) = 1.9x104 kg*m/s

Collisions

North

mtotal = 6.20x104 kg

ptotal = 1.9x104 kg*m/s

p = mv

v = p/m

v = (1.9x104 kg*m/s) / (6.20x104 kg) = 0.31 m/s north

Collisions

A 20.0-kg ball is moving north at 4.00 m/s toward a 5.00-kg ball that is stationary. After they collide, the 20.0-kg ball is moving at 2.40 m/s north. What is the 5.00-kg ball’s velocity after

the collision?Neglect friction.

CollisionsBefore Collision

m2 = 5.00 kg

North

m1 = 20.0 kg

4.00 m/s4.00 m/s

p1 = m1v1 p2 = m2v2

p1 = (20.0 kg)(4.00 m/s) p2 = (5.00 kg)(0 m/s)

p1 = 80.0 kg*m/s p2 = 0 kg*m/s

ptotal = (80.0 kg*m/s) + (0 kg*m/s) = 80.0 kg*m/s

CollisionsAfter Collision

m2 = 5.00 kg

North

m1 = 20.0 kg

2.40 m/s2.40 m/s ??????

ptotal = 80.0 kg*m/s

p1 = (20.0 kg)(2.40 m/s)

p1 = 48.0 kg*m/s

p2 = (80.0 kg*m/s) – (48.0 kg*m/s) = 32.0 kg*m/s

v2 = p2 / m2

v2 = (32.0 kg*m/s) / (5.00 kg) = 6.40 m/s

Collisions

A 0.0085-kg bullet strikes a stationary wooden block with a mass of 40.0 kg on a frictionless surface. After the impact, the block-bullet combo is moving at 0.20 m/s. What was the bullet’s original

velocity?

CollisionsBefore Collision

40.0 kg

m1 = 0.0085 kg

p1 = m1v1

Don’t know!

p2 = m2v2

p2 = (40.0 kg)(0 m/s)

p2 = 0 kg*m/s

??????

ptotal = Don’t know yet!

CollisionsAfter Collision

mtotal = 40.0085 kg = 40.0 kg

0.20 m/s0.20 m/s

ptotal = mtotalv = (40.0 kg)(0.20 m/s) = 8.0 kg*m/s

CollisionsBefore Collision

40.0 kg

m1 = 0.0085 kg

ptotal = 8.0 kg*m/s

??????

v1 = p1 / m1

v1 = (8.0 kg*m/s) / (0.0085 kg)

v1 = 940 m/s

Recoil (Collision in Reverse)

Whenever one object pushes another object away, momentum is still conserved.Think of it like a collision in reverse.

Recoil

A 60.-kg student on ice skates stands at rest on a frictionless frozen pond holding a 10.-kg brick. He throws the brick east with a speed of 18 m/s. What is the resulting velocity of the

student?

RecoilBefore Throwing

m1 = 60. kg m2 = 10. kg

ptotal = 0 kg*m/s

Recoil

ptotal = 0 kg*m/sAfter Throwing

m1 = 60. kg m2 = 10. kg

18 m/s???

p2 = m2v2

p2 = (10. kg)(18 m/s)

p2 = 180 kg*m/s

p1 + p2 = ptotal

p1 + (180 kg*m/s) = 0 kg*m/s

p1 = -180 kg*m/s

v1 = p1 / m1 = (-180 kg*m/s) / (60. kg)

v1 = -3.0 m/s