Physics 101 Quiz #5 SolutionOctober 16, 2002 Name

20 Minutes. Box your answers.

After an exhausting day ice fishing, a man is too tired to walk from his hole in thelake to the shore. He sits on a sled on the ice (it is frictionless), pulls out a machinegun (don’t leave home without one) and fires 100 bullets, horizontally, all in the samedirection.

1. [1 point] If the mass of a single bullet is 12.5 g (0.0125 kg) and its velocity is800 m/s, what is the momentum of a single bullet?

p = mv = 10kg m

s

2. [2 points] If the total mass of the man, the sled, and the machine gun is 100 kg,what is his velocity after firing 100 bullets? (You may neglect the mass lost in thegun due to the release of bullets.)

Let subscript 1 = man, subscript 2 = a single bullet.

pf1 + 100pf2 = p10 + 100p20

m1vf1 + 100m2vf2 = 0 + 0

vf1 = −100m2

m1vf2 =

(100)(0.0125)100

800ms

= −10ms

The negative sign indicates the man is moving opposite to the direction of thebullets.

Continued on the other side....

Rewrite and sign the honor pledge: “I pledge my honor that I have not violated theHonor Code during this examination.”

Signature

The man slides along for a while and then suddenly collides with a 400 kg moose,which was standing still on the ice. After the collision, the moose slides away atangle θ = 30◦, as shown in the diagram, while the man (still on his sled, carrying hismachine gun) ends up traveling at a 90◦ angle to his original path.

3. [2 points] What is the magnitude of the final velocity of the moose?

Use the x components of momentum. Initially, only the man has momentum inthe x direction. Finally, only the moose has momentum in the x direction. Ifthe moose’s velocity is vf2, the x component is vf2x = vf2 cos θ. Conservation ofmomentum gives:

m2vf2 cos θ = m1v01 ⇒ vf2 =m1v01

m2 cos θ=

100 · 10400 · cos 30◦

ms

= 2.88ms

4. [3 points] What is the final velocity of the man?

Use the y components of momentum. Initially there is no momentum in the ydirection, so the final y momentum must also be zero. Writing the man’s finalvelocity as vf1 and noting that the y component of the moose’s final velocity isvf2y = vf2 sin θ, conservation of momentum gives:

m1vf1 +m2vf2 sin θ = 0 ⇒ vf1 = −m2vf2 sin θm1

= −400 · 2.88 · sin 30◦

100ms

= −5.77ms

5. [2 points] Was the collision elastic? (Justify your answer quantitatively.)

KE0 = 12m1v

201 = 1

2 (100 kg)(10 m/s)2 = 5000 J

KEf = KEf1 + KEf2 = 12m1v

2f1 + 1

2m2v2f2

= 12 (100 kg)(5.77 m/s)2 + 1

2 (400 kg)(2.88 m/s)2 = 1660 J + 1660 J = 3320 J

No. The final mechanical energy is less than the initial mechanical energy. Energyis not conserved, so the collision is not elastic.

Physics 101 Quiz #5 SolutionOctober 15, 2004 Name

20 Minutes. Box your answers.

Problem 1 Two 20 kg sleds are standing at rest on a frozen lake close to each other.A 3 kg cat jumps from one sled to the other. The cat’s speed relative to the ice is 10m/s. In the following ignore any friction between the sleds and the ice!

(a)[2 point] What is the speed of the first sled after the cat jumped o!?

mass of the sled: M = 20 kgmass of the cat: m = 3 kg

pi = 0 kgms

= mcat · vcat + msled · vsled1 = pf

! vsled1 = " mcat

msled1· vcat = "1.5

ms

(b)[2 point] What is the speed of the second sled after the cat arrived there?

mcat · vcat = (mcat + Msled2) · vsled2

! vsled2 =mcat

mcat + Msled2· vcat = 1.3

ms

(c)[1 point] What is the velocity of the center of mass (of the 2 sleds and the cat)after the cat arrived on the second sled?

The velocity of the center of mass is 0 m/s as no external forces apply.

Continued on the other side....

(d)[5 points] The cat is still sitting on the second sled when the sled collides head-onwith a third sled which is at rest on the ice. The mass of the third sled is 20 kg. Whatis the velocity of the third sled after the elastic collision? If you do not trust yourresult for the velocity of the second sled that you derived in part (b), use v2 = 2 m/s.

m2 = 23m/s: mass of sled 2 with cat (m2 = m + M),m3 = 20m/s: mass of sled 3,vi2 = vsled2 = 1.3m/s: velocity of sled 2 before the elastic collisionvf2: velocity of sled 2 after the collision,vf3: final velocity of sled 3 after the collision.momentum conservation:

m2 · vi2 = m2 · vf2 + m3 · vf3

energy conservation:

12· m2 · v2

i2 =12· m2 · v2

f2 +12· m3 · v2

f3

From momentum conservation one gets

vf2 = vi2 "m3

m2· vf3

and putting this into the equation for energy conservation

12· m2 · v2

i2 =12· m2 ·

!vi2 "

m3

m2· vf3

"2

+12· m3 · v2

f3

12· m2 · v2

i2 =12· m2 · v2

i2 " m3 · vi2 · vf3 +12· m2

3

m2· v2

f3 +12· m3 · v2

f3

m3 · vi2 · vf3 = +12· m3

!1 +

m3

m2

"· v2

f3

vf3 = 2 · m2

m2 + m3· vi2

Using that in the momentum conservation we also can calculate the velocity of thesecond sled with the cat

vf2 = vi2 "m3

m2· vf3 = vi2 "

m3

m2· 2 · m2

m2 + m3· vi2 =

m2 " m3

m2 + m3· vi2

With the numbers plugged in:vf3 = 1.4m

svf2 = 0.09m

s

Rewrite and sign the honor pledge: “I pledge my honor that I have notviolated the Honor Code during this examination.”

Signature

PRINT YOUR NAME:

QUIZ 5 Physics 101, Fall 2005

SOLUTIONS

1.) A 100 kg man is training his 40 kg son to be a hockey goalkeeper. They are standing5 m apart on an icy surface (no friction) when the man whacks a 5 kg puck (initially at rest)with an impulse of 150 N · s toward his son.

a.) Considering the man+boy+puck system, what happens to the center of mass after thepuck is hit? Does it move toward the boy, toward the man, or remain stationary? Explainyour answer. [1 pt]

Neglecting friction, the net external force on the man+boy+puck system is zero, so the centerof mass remains stationary.

b.) What is the speed of the puck right after the man hits it? [2 pts]

J = !p = mpuckvpuck

vpuck = J/mpuck = (150 N · s)/(5 kg) = 30 m/s

c.) Find the speed of the boy after he catches the puck (and doesn’t drop it). [2 pts]

pf = pi ! (mboy + mpuck)vf = mpuckvpuck

vf = mpuckmboy+mpuck

vpuck = 3.3 m/s

QUIZ 5 Physics 101

2.) A pendulum bob of mass M = 1 kg is sus-pended from a rigid rod of length L = 10 cmand negligible mass. A bullet with mass m = 8 gtravelling at 300 m/s passes through the bob andemerges with speed v < 300 m/s. (Throughoutthis problem, ignore the fact that the bob has ahole in it after the bullet passes through!)

L

v = 300 m/s

m M

a.) Assuming that the bob has enough energy to just barely swing through a complete circle,find the initial speed of the bob just after the bullet exits. (Hint: assume v = 0 for the bobwhen it’s at the top of the loop.) [2 pts]

The rod is sti", so it can support the weight of the bob at the top of the loop. Therefore, thespeed at the top is nearly zero if the bob just has enough energy to complete a full circle (theforce of the rod on the bob balances the force of gravity).

KEf " KEi = PEi " PEf

0 " 1/2Mv2 = 0 " Mg(2L)

v =#

4gL = 2.0 m/s

b.) What is the final speed of the bullet? (Assume vbob = 1 m/s if you could not solve a.)[2 pts]

pf = pi ! mvbullet + Mvbob = mv0

vbullet = mv0−Mvbobm = (0.008 kg)(300 m/s)−(1 kg)(2m/s)

(0.008 kg) = 50 m/s

c.) Is this collision elastic? Justify your answer quantitatively. [1 pt]

KEi = 1/2mv20 = 1/2(0.008 kg)(300 m/s)2 = 360 J

KEf = 1/2mv2bullet + 1/2Mv2

bob = 1/2(0.008 kg)(50 m/s)2 + 1/2(1 kg)(2 m/s)2 = 12 J

Kinetic energy is lost, so this collision is inelastic.