Collisions

A + B → C+D+…

Conservation of Momentum

� Momentum in an isolated system in which a collision occurs is conservedconserved

� An isolated system will not have external forces

� Specifically, the total momentum before the collision will equal the total momentum after the collision

Conservation of Momentum, cont.

� Mathematically:

� Can be generalized to any

1 1 2 2 1 1 2 2i i f fm m m m+ = +v v v vr r r r

� Can be generalized to any number of objects

� Components of Momentum

� External forces? (short collision time)

Notes About A System

� Remember conservation of momentum applies to the system

� You must define the isolated � You must define the isolated system

So its not the velocity which is conserved but the momentum mv.

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Two objects collide head on. Their masses and initial velocities are given. If the 10 kg mass has a final velocity of -0.1 m/s, what is the final velocity of the 1 kg mass?

10 kg

1 m/s

1 kg

-1 m/s

10 kg

-0.1 m/s

1 kg

? m/s

initial final

Two objects collide head on. Their masses and initial velocities are given. If the 10 kg mass has a final velocity of -0.1 m/s, what is the final velocity of the 1 kg mass?

10 kg

1 m/s

1 kg

-1 m/s

10 kg

-0.1 m/s

1 kg

? m/s

A. -10 m/s

B. -1 m/s

C. 0.1 m/s

D. 10 m/s

Why do the same number of balls move on the other side?

A. Conservation of momentum

B. Conservation of energyenergy

C. Newton’s ghost

D. All answers are correct (except maybe 3).

Types of Collisions

� Momentum is conserved in any collision

� Inelastic collisions

� Kinetic energy is not conservedKinetic energy is not conserved

� Some of the kinetic energy is converted into other types of energy such as heat, sound, work to permanently deform an object (momentum is still conserved)

� Perfectly inelastic collisions occur when the objects stick together (momentum is still conserved)

� Not all of the KE is necessarily lost

Types of Collisions

� Elastic collision

� both momentum and kinetic energy are conservedare conserved

� Actual collisions

� Most collisions fall between elastic and perfectly inelastic collisions

73. A tennis ball of mass 57.0 g is held just above a basketball of mass 590 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m, as shown in Figure P6.69. (a) Find the magnitude of the downward velocity with which the basketball reaches the ground. (b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. (b) To what height does the tennis ball rebound?

1.2 m 1.2 m

vvT

vB

With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m. What are the velocities of the tennis ball and basketball when they strike the floor?

A. vT = 4.85 m/s, vB= -4.85 m/s

B. v = -4.85 m/s, v = -4.85 m/sB. vT = -4.85 m/s, vB= -4.85 m/s

C. vT = -4.85 m/s, vB= 4.85 m/s

D. vT = 4.85 m/s, vB= 4.85 m/s

1.2 m 1.2 m

v1.2 m

vvT

vB

vT

vB'

m1v1i + m2v2i = m1v1f + m2v2f

m1 = mB, v1i = vB׀

m2 = mT, v2i = vT

Elastic collision:

(1/2)m1v1i2 + (1/2)m2v2i

2 = (1/2)m1v1f2 + (1/2)m2v2f

2

Only for 1 dimensional elastic collisions

m1v1i + m2v2i = m1v1f + m2v2f

=> m1v1i - m1v1f = -m2v2i + m2v2f

=> m1(v1i - v1f) = m2 (v2f - v2i)

Only for 1 dimensional elastic collisions

Elastic collision:

(1/2)m1v1i2 + (1/2)m2v2i

2 = (1/2)m1v1f2 + (1/2)m2v2f

2

=> m1v1i2 - m1v1f

2 = -m2v2i2 + m2v2f

2

=> m1(v1i2 - v1f

2) = m2(v2f2 - v2i

2)

x2 – y2 = (x – y) (x + y)

=> m1 (v1i - v1f) (v1i + v1f) = m2 (v2f - v2i)(v2f + v2i)

Because m1(v1i - v1f) = m2 (v2f - v2i)

Only for 1 dimensional elastic collisions

v1i + v1f = v2i + v2f

or

)vv(vv f2f1i2i1 −−=−

or

More About Elastic Collisions (1-d)

� Both momentum and kinetic energy are conserved

� Typically have two unknowns (1d)

ffii vmvmvmvm 22112211 +=+ ffii vmvmvmvm 22112211 +=+

fifi vvvv 2211 +=+ 222

211

222

211 2

1

2

1

2

1

2

1ffii vmvmvmvm +=+

v1I v2Iv1F v2F

before after

Solve the equations simultaneously

Inelastic collisions

Kinetic energy is not conservedconserved

Momentum is still conserved

A gun at rest of mass M shoots a bullet of mass m. If velocity of the bullet is +vb what is the velocity of the gun in terms of M, m and vb

A. -(m/M)vb(m/M)vB. (m/M)vb

C. -(M/m)vbD. (M/m)vb

What are ∆p and ∆KE in terms of m, M, and vb?

A. 0 and -vb2(1+M/m)

B. 0 and (1/2)mv 2(1+m/M)B. 0 and (1/2)mvb2(1+m/M)

C. -mvb and (vbM/m)2

D. -mvb and (vbm/M)2

KEi = 0

KEf = (1/2)mvb2 + (1/2)M(m/M)2vb

2

pf Fp

pi

∆∆∆∆p = pf – pi = F∆∆∆∆t

How can the stranded astronaut reach the shuttle?

A. Kick her arms and legs like a swimmerswimmer

B. Throw the wrench towards the shuttle

C. Throw the wrench away from the shuttle

Rocket Propulsion

)ln(f

ieif M

Mvvv =−

Glancing Collisions

• For a general collision of two objects in three-dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is momentum of the system in each direction is conserved�

� Use subscripts for identifying the object, initial and final velocities, and components

fy22fy11iy22iy11

fx22fx11ix22ix11

vmvmvmvm

andvmvmvmvm

+=++=+

Glancing Collisions

• The “after” velocities have x and y components

• Momentum is conserved in the x direction and in the y direction

• Apply conservation of momentum separately to each direction