Phys 105 Second Midterm Examination Page 1 Saturday, 10-December-2011

PROBLEM 1 (25 points)

A block of mass m = 10 kg is attached to the free end of a

massless spring (k = 70 N/m) that has its equilibrium length, as

shown in the figure. A horizontal force FP = 200 N is applied to

the block which is initially at rest. The force FP pushes the block

2.0 m up along the rough incline with µk = 0.10 and θ = 37°.

(g = 10 m/s2, sin 37° = 0.60, cos 37° = 0.80)

gm

k

!

µkFP

(a) (10 pts) During the 2.0-m displacement, find the work done on the block by: (i) the force FP, (ii) the gravitational force,

(iii) the normal force, (iv) friction force, and (v) the spring force. (i)

!

WP = work done by ! F P

!

WP =! F P "! d = FPd cos 37° = (200 N)(2.0 m)(0.80)

!

WP = 320 J

(ii) WG = work done by force of gravity m! g

!

WG = m! g "! d = mgd cos(90° + 37°)

!

WG = (10 kg)(10 m/s2)(2.0 m)("0.60) --->

!

WG = "120 J

(iii)

!

WN = work done by normal force ! F N

!

WN =! F N "! d = FNd cos 90° --->

!

WN = 0

(iv)

!

Wfr = work done by friction force ! F fr

!

Wfr =! F fr "! d = Ffrd cos180° = #µkFNd

!

FN = (mg cos 37° + FP sin 37°) = 80 N +120 N = 200 N

!

Wfr = "(0.10)(200 N)(2.0 m) --->

!

Wfr = "40 J

(v)

!

WS = work done by spring force FS = "kx

!

WS = " kxdx0

d# = "

12kd 2 = " 1

2(70 N/m)(2.0 m)2

!

WS = "140 J

(b) (5 pts) Using energy method, find the speed of the block after the 2.0-m displacement.

!

Wnet = "K = K # K0 =12mv 2 # 0

!

v = 2Wnetm

=2(WP +WG +WN +Wfr +WS)

m

!

v = 2(20 J)10 kg

--->

!

v = 2.0 m/s

(c) (5 pts) Find the change in the internal energy of the block-spring-incline system during the 2.0-m displacement.

!

"Einternal = #Wfr

!

"Einternal = #(#40 J) --->

!

"Einternal = 40 J

(d) (5 pts) Find the power delivered to the 10-kg block by FP at the moment the block has been displaced 2.0 m.

!

P =! F " ! v = Fv cos#

where

!

F = FP = 200 N,v = 2.0 m/s,and " = 37°

!

P = 200 N( ) 2.0 m/s( ) 0.80( )

!

P = 320 W

Phys 105 Second Midterm Examination Page 2 Saturday, 10-December-2011

PROBLEM 2 (25 points)

Consider the track shown in the figure. The section PQ is a quarter-circle of

radius r and is frictionless. The horizontal section QR has a length L with a

coefficient of kinetic friction µk. The section RS under the spring is

frictionless. Block A of mass m is released from rest at P. After sliding on

the circular section it collides elastically with block B of mass 3m at rest at

Q. After the collision block B slides on the track and it compresses the

spring by a distance x. Treat each block as a particle. Express your answers

in terms of some or all of the given quantities and related constants as

needed.

g

k

m

3m

A

B

r

P

Q R Sµk

L

x

y

(a) (5 pts) Using energy method, find

!

! v A , the velocity of block A just before it collides with block B. Express your answer in

terms of unit vectors.

No friction and only gravitational force mg does work. So,

!

KP +UP = KQ +UQ

!

KP = 0, UP = mgr, KQ =12mvA

2 , choose UQ = 0

!

mgr = 12mvA

2 --->

!

vA = 2gr --->

!

! v A = 2gr( )ˆ i

(b) (5 pts) Find ! ! v B , the velocity of block B just after the collision. Express your answer in terms of unit vectors.

Elastic head-on collision ---> Relative velocities:

!

vA " vB = "( # v A " # v B) where vB = 0 --->

!

" v A = " v B # vA

Elastic head-on collision ---> Conservation of momentum:

!

mvA = m " v A + 3m " v B --->

!

vA = " v A + 3 " v B

!

vA = " v B # vA + 3 " v B --->

!

2vA = 4 " v B

!

" v B =2vA4

=vA2

=2gr2

=gr2

--->

!

" ! v B =

gr2

#

$ % %

&

' ( ( ˆ i

(c) (5 pts) Find the change in the mechanical energy of block B as it slides from Q to R.

!

"E = "K + "U ---> where

!

"U = 0 and "K =Wnet =Wfr

!

Wfr =! F fr "! d = Ffrd cos180° = #Ffrd = #µkFNL = #µk 3mgL

!

"E = #µk 3mgL

(d) (5 pts) Find the speed of block B just before it starts to compress the spring at R.

Work-energy principle:

!

Wnet = "K

!

Wnet =Wfr = "K = KR # KQ

!

"3µkmgL =123mvR

2 "123mvQ

2

!

"µkgL =12vR2 "

12vQ2 where

!

vQ = " v B =gr2

!

"µkgL =12vR2 "

gr4

!

vR2 =

gr2" 2µkgL = g

r2" 2µkL

#

$ %

&

' (

!

vR = gr2" 2µkL

#

$ %

&

' (

(e) (5 pts) Find the stiffness constant k of the spring.

No friction and only spring force does work. So,

!

Kf +Uf = Ki +Ui

where

!

Kf = 0, Uf =12kx2, Ki =

12

3mvR2 , Ui = 0

!

12kx2 = 1

23mvR

2 --->

!

k = 3mvR2

x 2

From part (d) above:

!

vR2 = g r

2" 2µkL

#

$ %

&

' (

!

k = 3mgx 2

r2" 2µkL

#

$ %

&

' (

Phys 105 Second Midterm Examination Page 3 Saturday, 10-December-2011

PROBLEM 3 (25 points)

A gun is fixed to the bottom end of a frictionless triangular block (of

inclination angle θ) which itself is pinned to a horizontal table. There

is a wooden block of mass m which is kept initially at rest, at a

height h from the surface of the table. A bullet of mass mB and

velocity

!

! v B leaves the gun in a direction parallel to the incline,

impacts the wooden block and becomes embedded. Just before the

impact the wooden block is released free from rest. Express your

answers in terms of some or all of the given quantities and related

constants as needed.

mmB

g

x

y

!

! v B

hGun

Woodenblock

Pins

(a) (10 pts) Find the maximum height hmax from the surface of the table the block+bullet system will reach.

Momentum is conserved:

momentum before = momentum after

!

mBvB = mB +m( )v --->

!

v = mBmB +m

vB

Total mechanical energy is conserved:

!

K1 +U1 = K2 +U2

Take U = 0 on the surface of the table.

!

12mB +m( )v 2 + mB +m( )gh = 0+ mB +m( )ghmax

!

12v 2 + gh = 0+ ghmax

!

hmax =12v 2 + h = 1

2gmB

mB +m"

# $

%

& ' vB

(

) *

+

, -

2

+ h

!

hmax =12g

mB2

mB +m( )2 vB

2 + h

(b) (5 pts) Find the work done by gravity on the wooden block+bullet system as it reaches to hmax after the instant of impact. Work done by gravity:

!

WG = "#U = #K

!

WG = K2 " K1 = 0"12mB +m( )v 2

!

WG = "12mB +m( )

mBmB +m

vB#

$ %

&

' (

2

!

WG = "12

mB2

mB +m( )vB2

(c) (10 pts) Suppose that the pins fixing the triangular block and the wooden block of mass m are removed. The triangular

block becomes free to slide horizontally on the table with no friction. If the gun is fired, find the speed of the gun+triangular

block system just after the bullet leaves the gun. Assume that the gun+triangular block system has a total mass mS.

Momentum is conserved in x direction:

!

! P initial, x =

! P final, x where

!

! P initial, x = 0

!

! P final, x = mBvB cos" ˆ i +mS

! v S = 0

!

! v S = "mBmS

vB cos# ˆ i

!

vS =! v S =

mBmS

vB cos" (speed)

Phys 105 Second Midterm Examination Page 4 Friday, 29-July, 2011

PROBLEM (4) (25 points)

A pendulum consists of a mass M hanging at the bottom end

of a massless rod of length L, which has a frictionless pivot at

its top end. A mass m moving in the horizontal direction with

velocity

!

! v hits M, goes into it and stays inside. Express your

answers in terms of some or all of the given quantities and

related constants as needed.

(a) (8 pts) Find the smallest value of v sufficient to cause the

pendulum (with mass m inside) to swing to the horizontal

position at which it momentarily comes to rest.

! v

L

g

x

y

mM

Conservation of momentum during the collision:

!

pi = pf

!

m1v1i +m2v2i = m1v1 f +m2v2 f

where

!

v2i = 0 ,

!

m1 = m ,

!

v1 f = v2 f = V ,

!

m2 = m +M

!

mv = (M +m)V

Conservation of mechanical energy after the collision:

!

Ki +Ui = K f +U f

where

!

Ui = 0 ,

!

K f = 0 --->

!

Ki =U f

!

12(M +m)V 2 = (M +m)gL --->

!

V = 2gL

!

mv = (M +m)V = (M +m) 2gL

!

v =(M +m) 2gL

m

(b) (8 pts) Find

!

! I , the impulse exerted on the pendulum by the mass m during the collision. Express your answer in unit

vectors.

Impulse-momentum theorem: The change in the momentum

of an object is equal to the impulse on the object.

Impulse exerted on M by m:

!

! I = "! p = ! p f #

! p i = MV" i # 0 = MV

" i

!

! I = M m

M +m"

# $

%

& ' v" i

(c) (9 pts) Assume now that the mass

!

m = M2

and the collision between m and M is elastic. After the collision M swings and

comes momentarily to rest at a height h above its initial position. Find this height h.

Elastic collision between m1 and m2 with v2i = 0:

!

v1 f =m1 "m2m1 +m2

v1i and

!

v2 f =2m1

m1 +m2v1i

where

!

m1 = m =M2

,

!

m2 = M ,

!

v1i = v

!

v1 f =

M2"M

M2

+Mv = " v

3 and

!

v2 f =2(M2)

M2

+Mv = 2v

3

Conservation of mechanical energy after the collision:

!

Ki +Ui = K f +U f

where

!

Ui = 0 ,

!

K f = 0 --->

!

Ki =U f

!

12Mv2 f

2 = Mgh --->

!

h =v2 f2

2g=12g(2v3)2 = 4v

2

18g

!

h = 2v2

9g

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