EXTRA QUESTIONS WEEK 5:

Question 1(a)

A particle A of mass m moving with a velocity of u collides elastically with another particle B of mass M, initially at rest. After collision, A and B move forward with velocities v and V respectively.

(a) Find v in terms of u and V.

By conserving the momentum right before and after the collision,

mu + M(0) = mν + MV mν = mu – MV ν = (mu-MV)/ m = u – (M/m)V M/m = (u-ν)/V--------------------------------------------------------------(1)

Assuming that the collision is perfectly elastic,

Ki = Kf

(1/2)mu2 + (1/2)M(0)2 = (1/2)mν2 + (1/2)MV2

mu2 = mν2 + MV2

m(u2 - ν2) = MV2

M/m = (u2 - ν2)/V2 -------------------------------------------(2)

By equating (1) and (2),

(u-ν)/V = (u2 - ν2)/V2

V2/V = (u2 - ν2)/(u-ν)

Since u2 - ν2 = (u+ν)(u-ν),

V = u+ν

ν = V-u #

Question 1(b)

Find the fractional kinetic energy lost by A in terms of m and M.

ΔKi/Ki = (Kf-Ki)/Ki

= (1/2)m( ν 2 -u 2 ) (1/2)mu2

= (ν2-u2)/u2

= (ν/u)2-1---------------------------------------------------------(3)

From equation (1),

M/m = (u-ν)/V

Since, ν = V-u V = u+ν Hence, M/m = (u-ν)/(u+ν)Mu+Mν = mu-mνν (M+m) = u(m-M)

ν/u = (m-M)/(M+m)--------------------------------------------------(4)

by substituting equation (4) into (3),

ΔKi/Ki = [(m-M)/(M+m)]2 -1 = (m-M) 2 -(M+m) 2 (M+m)2

= -4Mm/(m+M)2

-ΔKi/Ki = 4Mm/(m+M)2

#

Question 1(c)

What is the value of the fractional kinetic energy lost if M = 10m?

Given that M =10m,

ΔKi/Ki = -4(10m)m/(m+10m)2

= -40m2/121m2

= -0.331 #

Question 2

A vessel at rest at the origin of an xy coordinate system explodes into 3 pieces. Just after explosion, one piece of mass, m moves with velocity –30 m/s in the x direction while the second piece which is also of mass, m moves with the same velocity as the first piece but along the y direction. Determine the magnitude and the direction of the third piece which has a mass of 3m.

y v = ? -30m/s

x

-30m/s

By conserving the momentum along the x-axis,

-30m + 3mvx = 0 vx = 10 m/s

By conserving the momentum along the y-axis,

m

m

3m

-30m + 3mvy = 0 Vy = 10 m/s

v = √( vx 2 + Vy

2) = √200 = 14 m/s #

tan θ = Vy/vx

= 1 θ = 45o

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Question 3

Two 500 g blocks of wood are 2.0 m apart on a frictionless table. A 10 g bullet is fired at 400 m/s towards the blocks. It passes all the way through the first block, and then embeds itself in the second block. The speed of the first block immediately afterwards is 6.0 m/s. What is the speed of the second block after the bullet stops?

By conserving the momentum for block 1,

400(10) + 500(0) = 500(6) + 10v

Whereby v is the speed of the bullet after passing block 1

V = 100 m/s

By conserving the momentum for block 2,

500(0) + 10(100) = (500 + 10)vf

Whereby vf is the common speed of the bullet and block 2 after collision

vf = 1.96 m/s #

Question 4

A wooden block is cut into two pieces, one with three times the mass of the other. A depression is made in both faces of the cut, so that a firecracker can be placed in it with the block reassembled. The reassembled block is set on a rough-surfaced table, and the fuse is lit. When the firecracker explodes, the two blocks separate and slide apart. What is the ratio of distances each block travels?

Let the mass of one of the two pieces of block s be m and the other be 3m,

By conserving the momentum right before and after the collision,

mvlight – 3mvheavy = 0vlight = 3vheavy ------------------------------(1)

by using the work-kinetic theorem equation for block 1,

Wf = Kf – Ki

Ki = 0Wf =Kf

Ff.slight.cos θ = ½ mvlight2

-Ff. slight = ½ mvlight2

slight = - ½ mvlight2

Ff

Similarly for block 2,

Wf = Kf – Ki

Ki = 0Wf =Kf

Ff.sheavy.cos θ = ½ mvheavy2

-Ff. sheavy = ½ mvheavy2

Sheavy = - ½ mvheavy2

Ff

Sheavy/slight = vheavy2/vlight

2 ----------------------------------------(2)

By substituting (2) into (1),

Sheavy/slight = vheavy2/9 vheavy

2

= 1/9 #

Question 5(a)

Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings downward and strikes a 4.60 kg ball that is at rest, as the figure shows.

(a) Using the principle of conservation of mechanical energy, find the speed of the 1.50 kg ball just before impact.

By conserving the energy f the 1.50 kg bob before being swung downward and right before collision,

Ki + Ui = Kf½mvf

2 = ½mvi

2+ mghvf = √(vi + 2gh) = √(25 + 2×9.81×0.300) = 5.56 m/s #

Question 5(b)

Assuming that the collision is elastic, find the velocities (magnitude and direction) of both balls just after the collision.

By conserving the momentum of the two balls before and after collision,

1.50(5.56) + 4.60(0) = 1.50v1 + 4.60 v2

V1 = 5.56 – 3.10v2 ------------------------------------------------------(1)

Because the collision is perfectly elastic,

Ki = Kf½(1.50)(5.56)2

= ½(1.50)v12+ ½(4.60)v2

2

v12 + 3.10 v2

2 = 30.9 ------------------------------------------------------(2)

by substituting (1) into (2),

(5.56 - 3.10v2)2 + 3.103.10v22 = 30.9

v2 = 2.71 m/s #

v1 = 5.56 – 3.10(2.71) = -2.84 m/s #

Question 5(c)

Ignoring air resistance, how high does each ball swing after the collision?

By conserving the energy of block 2 right after the collision,

Ki = Uf½mvi

2 = mgh

h = vi2/2g

= 2.712/(2×9.81) = 0.37 m #

Similarly for block 1,

H = 2.842/(2×9.81) = 0.411 m

#

Question 6

By accident, a large plate is dropped and breaks into three pieces. The pieces fly apart parallel to the floor. Figure above indicates the velocities and positions of the three broken pieces immediately after the impact with the floor. As the plate falls, its momentum has only a vertical component and no component parallel to the floor. After the collision, the component of the total momentum parallel to the floor remains zero. Using the data shown in figure above, find the masses m1 and m2.

V1x = 3sin(25o) = 1.27 m/sV1y = 3cos(25o) = 2.72 m/s

V2x = 1.79sin(45o) = 1.27 m/sV2y = 1.79cos(45o) = 1.27 m/s

V3 = V3y = 3.07 m/sV3x = 0 m/s

By conserving momentum in the x-axis,

-1.27m1 + 1.27m2 = 0m2 – m1 =0m2 = m1 = m

By conserving momentum in the y-axis,

2.72m + 1.27 m -1.30 (3.07) = 0m = 1.00 kg #

EXTRA QUESTIONS WEEK 6

Question 1(a)

A child sitting 1.10 m from the center of a merry-go-round moves with a speed of 1.25 m/s.(a) Calculate the centripetal acceleration of the child

ac = v2/r = 1.252/1.10 = 1.42 m/s2

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Question 1(b)

Calculate the net horizontal force exerted on the child (mass = 25.0 kg)

∑F = mac

= 25.0(1.42) = 35.5 N #

Question 2

A coin is placed 11.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin

remains fixed on the turntable until a rate of 36 rpm is reached and the coin slides off. What is the coefficient of static friction between the coin and the turntable?

Fs = Fc

μsFN = mac

μs = ac/g = v2/rg --------------------------------------(1)

ω = 36 rev × 1 min × 2п rad min 60 s 1 rev = 3.77 rad/s r = 0.11 m

v = rω

v2/r = (rω)2/r = rω2 --------------------------------------(2)

By substituting (2) into (1),

μs = rω2/g = (0.11 × 3.772)/9.81 = 0.16 #

Question 3

An object of mass 2.00 kg is attached to the end of a string of length 1.00 m and whirled in a vertical circle at a linear speed of 4.00 m/s. Find the tension in the string at the top and bottom of the circle.

vT = 4.00 m/sL = 1.00 mm = 2.00 kgac = v2/r = 4.002/1.00 = 16.0 m/s #

At the top,

T + mg = mac

T = m (ac – g) = 2.00 (16.0-9.81) = 12.4 N

At the bottom,

T – mg = mac

T = m (g + ac) = 2.00 (9.81 + 16.0) = 51.6 N #

Question 5(a)

×

T mg

× T

mg

A bucket of mass 2.00 kg is whirled in a vertical circle of radius 1.10 m. At the lowest point of its motion the tension in the rope supporting the bucket is 25.0 N.

(a) Find the speed of the bucket.

At the bottom,

T – mg = mac

= mv2/rv = √r(T/m – g) = √1.10(25.0/2.00 – 9.81) = 1.72 m/s #

Question 5(b)

How fast must the bucket move at the top of the circle so that the rope does not go slack?

At the top,

T + mg = mac

= mv2/rT = mv2/r – mg

If the string does not go slack,

mv2/r ≥ mg

the minimum speed at which the string remains taut(stretched),

T = 0 N

mv2/r – mg = 0v2/r = gv = √rg =√1.10×9.81 = 3.28 m/s

#

Question 6(a)

An airplane flies at a constant speed of 180 ms -1 in a horizontal circle of radius 20 km. A pendulum bob is hanging from the ceiling of the plane cabin by using a string which makes an angle of θ to the vertical.

(a) Find the acceleration of the airplane.

ac = v2/r = 1802/(20 × 1000) = 1.62 m/s #

Question 6(b)

Name the force on the pendulum bob. Draw and label the diagram -Tsinθ

θ

Calculate the value of the angle θ.

Because the plane moves in a horizontal circle,

T

Tcosθ

mg

∑Fx = mac

-Tsinθ = -mac

Tsinθ = mac ----------------------------------------------------------------------------------------(1)

∑Fy = 0Tcosθ – mg = 0Tcosθ = mg --------------------------------------------------------------(2)

(1)÷(2),

tanθ = ac /g

= v2/rg = 1802/(20000×9.81) = 0.165θ = 9.37o

=9o 22’ #

Question 7

A block (mass = 2.0 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1.1 × 10-3 kg m2), as the figure shows. Initially the pulley is prevented from rotating and the block is stationary. Then, the pulley is allowed to rotate as the block falls. The cord does not slip relative to the pulley as the block falls. Assume that the radius of the cord around the pulley remains constant at a value of 0.040 m during the block’s descent. Find the angular acceleration of the pulley and the tension in the cord.

I = 1.1 × 10-3 kgm2

ωo = 0r = 0.040 mm = 2.0 kg

∑Fy = ma

T – mg = - ma

a = g – T/m ---------------------------------------------------------(1)

∑τ = Iα

Assuming that the pulley is frictionless,

Tr = Iαα = Tr/I

a = rα = Tr2/I --------------------------------------------------------------(2)

By equating (1) and (2)

g – T/m = Tr2/I

T = mgI/(I + r2m) = (2.0 × 9.81 × 1.1×10-3)/( 1.1×10-3 + 2.0 × 0.042) = 5.02 N #

α = Tr/I = (5.02 × 0.04)/(1.1 × 10-3) = 183 rad/s2

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