Momentum & Collisions

Ball BounceWhich ball experiences the greatest change in its momentum (both have equal mass)?

• Green one• Red one• Both have the same change

Example

Initial conditions:

mm = 0.1kg, ms = 1000kg,

vmi = 1000m/s, vsi= 0

Momentum before collision:Pti = mmvmi + msvsi = 0.1(1000) + 1000(0) = 100 Ns

Final conditions: There is only one object after the collision (combined mass of 1000.1kg)Momentum after collision:

Ptf = m(m+s)v(m+s)f = 1000.1(v(m+s)f )

Meteor strikes a spacecraft and becomes embedded.

Momentum Conservation says: Pti = Ptf

100 = 1000.1(v(m+s)f ) therefore v(m+s)f = 0.1 m/s

Same Initial conditions as before:

mm = 0.1kg, ms = 1000kg,

vmi = 1000m/s, vsi= 0

Momentum before collision:Pti = mmvmi + msvsi = 0.1(1000) + 1000(0) = 100 Ns

Final conditions:

mm = 0.1kg, ms = 1000kg,

vmf = -1000m/s, vsf = ?

Momentum after collision:Ptf = mmvmf + msvsf = 0.1(-1000) + 1000(vsf)

Meteor strikes a spacecraft and rebounds.

Momentum Conservation says: Pti = Ptf

100 = -100 +1000 (vsf) therefore vsf = 0.2 m/s

Twice as much as if the meteor sticks!

0.2 m/sec

Example

Skateboarder

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rP ti =

r P tf

mp +s+b

r v ( p +s+b )i = mp +s

r v ( p +s) f + mb

r v bf

r v bf =

1

mb

mp +s+b

r v ( p +s+b )i − mp +s

r v ( p +s) f( )

=1

686(4) − 80(5)( )

= −9.33 m/s

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rJ b =

v F bΔt = mb

r v bf − mb

r v bi

= 6(−9.33) − 6(4)

= −80Ns

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rJ p +s =

v F p +sΔt = mp +s

r v ( p +s) f − mp +s

r v ( p +s)i

= 80(5) − 80(4)

= 80Ns

A 75 kg person on a 5 kg skateboard has a velocity of vx = 4 m/s. Theperson is holding a 6 kg ball, which he throws. After he throws the ball, he ismoving at vx = 5 m/s. How fast (relative to the ground) did he throw the ball andin what direction? What was the impulse on the ball? How did it compare to theimpulse on the person+skateboard system?

Space Probe

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rP ti =

r P tf

mpi

r v pi = mpf

r v pf + me

r v ef

r v pf =

1

mpf

mpi

r v pi − me

r v ef( )

=1

62806360(130) − 80(−248)( )

A 6360 kg space probe moving nose-first toward Mars at 130 m/s relative to the Sun fires its rocket engine, ejecting 80.0 kg of exhaust at a speed of 248 m/s relative to the space probe. What is the final velocity of the probe?

Here we assume that the engine fires in the opposite direction as the initial motion of the probe

Suppose the probe fires its engine in a direction perpendicular to the initial direction of travel. What is its new velocity?

Space Probe

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rP tiy =

r P tfy

0 = Ppfy + me (vefy )

Ppfy = −me (vefy )

= −80(−248)

=19840 kg ms

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rP tix =

r P tfx

mpi

r v pix =

r P pfx +

r P efx

r P pfx = mpi

r v pix − me

r v efx

= 6360(130) − 80(130)

= 816400 kg ms

y

x

€

rv pfx =

r P pfx /mpf

= 816400 /6280 =130 ms

r v pfy =

r P pfy /mpf

=19840 /6280 = 3.16 ms

2D CollisionA 0.17 kg hockey puck on an ice rink moving at (20 m/s, 30 m/s) collides with a 20 kg curling stone moving at (-2 m/s, 0 m/s) and bounces off the stone, moving at (-20 m/s, 25 m/s). What is the final velocity of the curling stone?

€

rv s,i

€

rv p,i

€

rv p, f

+x

+y

€

rv s, f = ?

€

rP i =

r P f

r P i,x =

r P f ,x ,

r P i,y =

r P f ,y

mp

r v p,i,x + ms

r v s,i,x = mp

r v p, f ,x + ms

r v s, f ,x

r v s, f ,x =

1

ms

mp

r v p,i,x + ms

r v s,i,x − mp

r v p, f ,x( )

=1

20(0.17)(20) + (20)(−2) − (0.17)(−20)( )

= −1.66 ms

2D CollisionA 0.17 kg hockey puck on an ice rink moving at (20 m/s, 30 m/s) collides with a 20 kg curling stone moving at (-2 m/s, 0 m/s) and bounces off the stone, moving at (-20 m/s, 25 m/s). What is the final velocity of the curling stone?

€

rv s,i

€

rv p,i

€

rv p, f

+x

+y

€

rP i =

r P f

r P i,x =

r P f ,x ,

r P i,y =

r P f ,y

mp

r v p,i,y + ms

r v s,i,y = mp

r v p, f ,y + ms

r v s, f ,y

r v s, f ,y =

1

ms

mp

r v p,i,y + ms

r v s,i,y − mp

r v p, f ,y( )

=1

20(0.17)(30) + (20)(0) − (0.17)(25)( )

= 0.04 ms

€

rv s, f = (−1.66 , 0.04) m

s

€

rv s, f = ?

Elastic and inelastic collisionsCollisions are classified by whether or not energy is conserved in the collision

If energy is conserved then the collisions are deemed to be elastic • Collisions between atoms are thought to be elastic since they don’t

really come into contact when they interact• Billiard ball collisions are also “nearly” elastic

If energy is not conserved then the collisions are called inelastic• Collisions between cars• Objects that stick together during the collision are said to be

completely inelastic

The collisions that you observe every day are to some degree inelastic, but many can be approximated as nearly elastic.

In both collisions, momentum is conserved (absence of external forces)

A 10 kg block moving to the right at 5 m/s collides elastically with a 4 kg boxmoving to the left at 2 m/s. What are their velocities immediately after the collision?

1D Collision

€

rv L,i

€

rv R ,i

€

rP i =

r P f

mL

r v L,i + mR

r v R ,i = mL

r v L, f + mR

r v R , f

r v L, f =

1

mL

mL

r v L,i + mR

r v R ,i − mR

r v R , f( )

Check on your own!

€

rv L, f =1 m

s

r v R , f = 8 m

s

€

KE i = KE f

12 mL vL,i( )

2+ 1

2 mR vR ,i( )2

= 12 mL vL, f( )

2

+ 12 mR

r v R , f( )

2

BargeA coal barge with a mass of 21,500 kg drifts along a river. It passes under a big coal hopper kept full of coal that funnels coal at a constant rate of 200 kg/s from a height of 2 m above the barge. When the barge passes under the hopper, it is loaded with 13,500 kg of coal. What is the speed of the unloaded barge if, after loading, the full barge has a speed of 0.5 m/s?

€

rP t,x,i =

r P t,x, f

mb

r v b,i = mb +c

r v (b +c ), f

21500r v b,i = 35000(0.5)

r v b,i = 0.814 m

s

What is the force that acted on the barge during the loading?

€

vF bΔt = Δ

r p b = mb

r v b, f − mb

r v b,i

v F b =

1

Δtmb

r v b, f − mb

r v b,i( )

Δt =13500kg

200 kgs

= 67.5s

v F b =

1

67.521500(0.5) − 21500(.814)( )

v F b = −100N

ParatrooperIn February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 56 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the magnitude of the force on him from the snow was at the survivable limit of 1.2 × 105 N. What are (a) the minimum depth of snow that would have stopped him safely and

(b) the magnitude of the impulse on him from the snow?

€

vF pΔt = m

r v f − m

r v i

v F pΔt = 85(0) − 85(56)v F pΔt = −4760Ns

€

W = ΔKE

Fpdcos θ( ) =1

2mv f

2 −1

2mv i

2

€

d =1

2Fp cos θ( )mv f

2 − mv i2

( )

= −85 562

( )

2(1.2 ×105)(−1)

=1.11 m