A bullet of mass m and velocity v is ﬁred into a wood block of A bullet of mass m and velocity v...

1

A bullet of mass m and velocity v is fired into a wood block of mass M intially at rest on a frictionless surface. The bullet buries itself into the block and then the block is seen to have a final velocity vf.

Was this an elastic collision? A) Yes B) No

True or False: ½ mv2 = ½ (M+m)vf2 A) True B) False

True or False: mv = Mvf A) True B) False

This is an inelastic collision because the bullet sticks to the block.

L30 M 11/3/14 a*er lecture

2





As an inelastic collision, KE is not conserved.


3





Even in an inelastic collision, momentum is conserved, but we must consider the entire mass of the bullet + block. L30 M 11/3/14 a*er lecture

L30 M 11/3/14 a*er lecture 4

Assignments

Today: •  Finish up our discussion of orbits and gravity, found in Ch. 8, as

well as linear momentum and its conservation (including impulse), found in Ch. 9.

•  Profs Radzihovsky’s and Munsat’s clicker questions are placed at the end of today’s lecture notes.

•  Will move on to rotational motion, found in Ch. 10.

Announcements: •  CAPA 10 and HW 10 are due this week. •  Clicker questions from Prof. Radzihovsky’s and Prof. Munsat’s

lectures last week will appear at the end of today’s Lecture Notes. •  You should read Ch. 10 now.


p = mv is conserved ⇒ ptotal = constantTwo objects collide: m1

v1i +m2v2i = m1

v1 f +m2v2 f

In 1D (scalar equation): m1v1i +m2v2i = m1v1 f +m2v2 f

If it is an "elastic collision", KE is also conserved:

12m1v1i

2 + 12m2v2i

2 = 12m1v1 f

2 + 12m2v2 f

2

If it is not elastic, it is inelastic. Total energy is conserved, KE is not.If it is a totally inelastic collision, then the objects stick together after the collision so that there is a single final speed vf : m1v1i +m2v2i = (m1 +m2 )vf

Linear Momentum (derived from NII and NIII)

(initial momentum = final momentum)


NII is written elegantly in terms of momentum:

Fnet =

ddtp

This can be productively applied to one particle. If the net force = constant:

FnetΔt = Δp

The change of momentum of a particle during a collision is referred to as the impulse.

Linear Momentum

(Units: [F][t] = N s) (impulse)

7

I= Δ p= FnetΔtA ball bounces off the floor as shown.

The direction of the impulse of the ball, , is A) Straight up,

B) Straight down, C) To the right,

D) To the left,

E) Indeterminate.

Δ p

v1

v2

θ θ

Note: the angle of incidence is equal to the angle of reflection.

p1 + Δ p

= p2


8

Comment on Elastic Collisions


9

Comment on Elastic Collisions

for elastic collisions

Derivation is in the book and in Prof. Dubson’s notes on Ch. 9.


10

m m

Consider a suspended blue ball that collides elastically with a suspended red ball.

(like Newton’s cradle)


vB

vA = 0

11

m m

Consider a suspended blue ball that collides elastically with a suspended red ball.

m m

vB − 0 = vB = −(0 − vA' ) = vA

'

After impact is the situation at right possible: A) Yes B) No


vB

vA' = vB

vB' = 0

vA = 0

12

G ≈ 6.67 x 10−11Nm2 / kg2

Newton’s Law of Universal Gravitation

F = G m1m2

r2


g(r) = GMr2

(1) Acceleration of gravity:

(2) Speed of a satellite in a circular orbit:

vcircular orbit =GMr

(3) Relation between period τ and radius r of a circular orbit:

τ 2 = 4π 2

GM⎛⎝⎜

⎞⎠⎟r3

(4) Saw how the mass of the earth and the sun can be computed from knowledge of their satellites’ orbits. (5) Also saw how planetary radii and periods can be inferred from knowledge of other planetary radii and periods.

13

Gravitational Potential energy


U(r) = −GmMr

F = − dUdr

= GmMr2

Note: we chose the reference level for potential energy such that U(r) < 0 everywhere, except at infinity where it is = 0.

14



U(r) = −GmMr

F = − dUdr

= GmMr2

Etot = KE + PE = 12mv2 −G mM

rTotal energy of an object of mass m moving at speed v subject only to the attractive force of gravity of an object of mass M a distance r away.

15 L30 M 11/3/14 a*er lecture

A projectile is fired with initial speed v0 at radius r0 from the center of an airless planet of mass M (r0 is significantly greater than the radius of the planet). The projectile rises to a maximum distance rmax and then falls back down to the ground, and achieves a final speed vfinal just before it hits the ground.

A student wishes to compute rmax. What is the correct Conservation of Energy equation, she should start with?

A) 12mv0

2 −G Mmr0

= −G Mmrmax

B) 12mv0

2 −G Mmr0

= 12mvfinal

2 −G Mmrmax

C) 12mv0

2 = −G Mmrmax

D) −G Mmr0

= 12mvfinal

2 −G Mmrmax

Initial point

Final point, v = 0

16



U(r) = −GmMr

Etot = 0

Etot = 0 Unbound Orbit Parabola v0 = vescape Etot < 0 Bound Orbit Circle, Ellipse v0 < vescape Etot > 0 Unbound Orbit Hyperbola v0 > vescape

F = − dUdr

= GmMr2

= 2GMRE

0 = Etot = KE + PE = 12mv2 −G mM

r2


Does escape speed depend on launch angle? That is, if a projectile is given an initial speed vo, is it more likely to escape an (airless) planet, if fired straight up than if fired at an angle? A)  Yes B)  No C)  Impossible to tell from the information given.

The speed of escape depends only on the relationship between the object’s KE and PE at launch, which are independent of angle.


The following clicker questions were used by Profs Radzihovsky and Munsat in their lectures on Wed and Friday, Oct 29 and 31. These are Lectures 28 and 29.

19

In which situation is the magnitude of the total momentum the largest? A)  Situation I. B)  Situation II. C)  Same in both.

mv + 0 = mv

mv + (2m)(-v) = -mv

Magnitudes are the same:


A car is sitting on the surface of the Earth and both the car and the Earth are at rest. (Pretend the Earth is not rotating or revolving around the Sun.) The car then accelerates to a final velocity.

After the car reaches its final velocity, the magnitude of the Earth’s momentum is ___________ the magnitude of the car’s momentum. A)  more than B)  the same as C)  less than D)  Cannot answer, need more information.

20

Total momentum of the car/Earth system is zero. So, as the car acquires momentum to the right, the Earth acquires the same momentum to the left.

| MEvE |=| Mcvc | → | vE |= Mc

ME

⎛⎝⎜

⎞⎠⎟

| vc || vc |


21

Two masses of size m and 3m are at rest on a frictionless table. A compressed, massless spring between the masses is suddenly allowed to uncompress, pushing the masses apart.

After the masses separate, the speed of m is _______ the speed of 3m. A) the same as B) twice C) 3 times D) 4 times.

ptot = 0 = mv1 + 3mv2

v1 = −3v2 → | v1 | = 3 | v2 |

v1 v2 1 2


22


After the masses separate, the KE of m is _______ the KE of 3m. A) the same as B) twice C) 3 times D) 9 times.

KE2 =12(3m)v2 = 3

2mv2

KE1 =12m(−3v)2 = 9

2mv2 = 3KE2

-3v v

Let v2 = v, then v1 = −3v1 2


23


While the spring is in contact with the masses, how does the magnitude of the force of m on 3m compare to the magnitude of the force of 3m on m. A) the same as B) less than C) greater than D) unknown

-3v v

By NIII

Let v2 = v, then v1 = −3v1 2


24

I= Δ p= FnetΔtA ball bounces off the floor as shown.

The direction of the impulse of the ball, , is A) Straight up,

B) Straight down, C) To the right,

D) To the left,

E) Indeterminate.

Δ p

p

1 = (px ,− py )

p

2 = (px , py )

Δp = p

2 − p

1 = (px , py ) − (px ,− py ) = (0,2py )

v1

v2

θ θ

px

py

-py

p1 + Δ p

= p2


25

Suppose a tennis ball and a bowling ball are rolling toward you. Both have the same momentum p and you exert the same force F to stop each. How do the time intervals to stop them compare? A)  It takes less time to stop the tennis ball.

B)  Both are the same.

C)  It takes less time to stop the bowling ball.

Δt = ΔpF

which is the same for both the tennis and bowling balls.

I= Δ p= FnetΔt


A fast-ball thrown at a batter has a momentum of magnitude |pi| = (0.3kg)(40m/s) = 12 kg m/s. The batter hits the ball back in a line drive with momentum of magnitude |pf| = (0.3kg)(80m/s) = 24 kg m/s. What is the magnitude of the impulse |Δp|? A)  12 kg m/s B)  24 kg m/s C)  36 kg m/s D)  0 kg m/s E)  None of these.

26

I= Δ p= FnetΔt

= 24 - (-12) kg m/s = 36 kg m/s

= 24 kg m/s = -12 kg m/s


27

A big ball of mass M = 10m and speed v strikes a small ball of mass m at rest. After the collision, could the big ball come to a complete stop and the small ball take off with speed 10 v? A) Yes this can occur. B) No, because it’d violate conservation of momentum. C) No, because it’d violate conservation of energy.

pi = (10m)v = 10mvpf = m(10v) = 10mv

KEi =12(10m)v2 = 5mv2

KEf =12m(10v)2 = 50mv2


A bullet of mass m and velocity v is ﬁred into a wood block of A bullet of mass m and velocity v...

Documents

Transcript of A bullet of mass m and velocity v is ﬁred into a wood block of A bullet of mass m and velocity v...