1

Solutions to PC1431 Tutorial 3

1(a) Refer to the diagram below,

By conservation of energy, taking the GPE of pendulum to be 0 at the reference line, we have

Einitial = Eafter hitting peg

mghi =1

2mv2 +mghf

If hi = 2 (L− d), then the pendulum will reach hf = 2 (L− d) at zero speed, so hi = hf . It will beso when the pendulum is released at hi < 2 (L− d).

Hence we have 2 situations: (i) pendulum reach hf at zero speed, or (ii) pendulum reach hf atnon-zero speed.

The conditions for situation (i) and (ii) will be hi < 2 (L− d) and hi > 2 (L− d) respectively.

If you released the pendulum below the peg, i.e. hi < L − d < 2 (L− d), then we have hi = hf .(shown)

1(b) By conservation of energy, just before the string hits the peg, the KE of the sphere is:

1

2mv2b = mgL

If the sphere were to swing a complete circle centered on the peg, radius, R = L−d, then the changein GPE is:

∆GPE = mg × 2 (L− d)

Since the sphere is making a vertical circular motion about the peg, at the top of the sphere, fromthe FBD, we have:

T +mg =mv2tR

At the top position, the total energy of the pendulum is:

E =1

2mv2t +mg2(L− d)

2

By conservation of energy, we have E = mgL

mgL =1

2mv2t +mg2(L− d)

v2t = 2gL− 4g(L− d) = 4gd− 2gL

For minimum d, vt will be minimum, so the centripetal force will be minimum, hence we requireT = 0, then

mg =mv2tR

v2t = gR = g (L− d) = 4gd− 2gL

gL− gd = 4gd− 2gL

d =3

5L (shown)

2(a) By work-energy theorem, energy lost due to friction is the decrease in the kinetic energy of theparticle.

WD by friction =1

2mv2i −

1

2mv2f =

1

2(0.400)

(82 − 62

)= 5.60 J

2(b) WD by friction = fkd = µkmgd

µk =∆KE

mgd=

5.60

(0.4) (9.8) (2π) (1.5)= 0.152

2(c) We will need to �nd total distance travelled by the particle.

fkD =1

2mv2i

µkmgD =1

2mv2i

D =v2i

2µkg=

82

2 (0.152) (9.8)= 21.5 m

The total number of revolutions is:

n =D

2πR=

21.5

2π (1.5)= 2.29

3 The collision happens at the bottom of the bowl when the 2 masses reach there. At the bottom ofthe bowl, there is no external forces acting along the direction of motion of the particles, hence wecan apply conservation of linear momentum. Before that, we need to �nd the speeds of the particle.

By conservation of energy, mg∆h = 12mv

2, we obtain

v =√

2g∆h

The speeds of the particles are the same at the bottom of the bowl. Since they are moving at rightangles to each other just before the collision, the sketch for the situation is:

3

Based on the coordinates system, the total momentum of the system before the collision is:

−→p i = m1−→u 1 +m2

−→u 2 = 2v i− 6v j

By conservation of linear momentum, the total momentum of the system after the collision is:

−→p f = 2v i− 6v j

Since the collision is completely inelastic, the masses will move up the bowl as one mass, M =m1 +m2, together. De�ne the �nal height reached as hf , by conservation of energy:

p2f2M

= Mghf

hf =p2f

2M2g=

(2v)2

+ (6v)2

2M2g=

40 (2g) (12)

2 (8)2g

= 7.5 cm

4 The sketch of the system is:

By conservation of linear momentum:

Mvi = m1v1 +m2v2

10(25) = 3v1 + 7v2

3v1 + 7v2 = 250 (1)

By conservation of energy,

1

2Mv2i + 1500 =

1

2m1v

21 +

1

2m2v

22

(10) (25)2

+ 3000 = 3v21 + 7v22

3v21 + 7v22 = 9250 (2)

From Eq. (1), we have v2 = 17 (250− 3v1). Substitute into (2),

3v21 + 7

[1

7(250− 3v1)

]2= 9250

30v21 − 1500v1 − 2250 = 0

v21 − 50− 75 = 0

v1 =50±

√502 + (75)

2

= 51.5ms−1 or − 1.46ms−1(N.A.)

Since the lighter fragment moves horizontally forward after explosion, we have v1 = 51.5ms−1

With that, we can �nd v2 = 13.7ms−1 from Eq. (1).

4

5. By symmetry, xCM = 0. Based on the diagram,

To �nd yCM , starting from the formula y = 1M

´y dm.

Length of the rod, L = R(π2

).

Based on the diagram, choosing a point on the arc, dm = λ ds

Mass per unit length, λ =M

L=

M

Rπ2

=2M

πR

Since it is a circular arc, we write ds = Rdθ, then dm = λ ds = 2MπRRdθ = 2M

π dθ

From the diagram, the y-coordinate of dm is y = R sin θ. The lower limit is θ = π4 , the upper limit

is θ = 3π4 .

yCM =1

M

ˆy dm =

1

M

ˆ 3π4

π4

R sin θ2M

πdθ

=2R

π

ˆ 3π4

π4

sin θ dθ

=2R

π[− cos θ]

3π4π4

=2R

π

[cos

π

4− cos

3π

4

]=

2√

2R

π

The distance x from the center of the rod is:

x = R− 2√

2R

π= 0.0997R = 0.0635L

5

6 Based on the diagram below:

From the free-body diagram,

mg cos θ − n = mv2

R

When the block loses contact, n = 0, so

mg cos θ = mv2

R

mgR cos θ = mv2

1

2mv2 =

1

2mgR cos θ

By conservation of energy,

mgx =1

2mv2

From the diagram, x = R−R cos θ. Then, we have

mgx =1

2mgR cos θ =

1

2mg (R− x)

3

2x =

1

2R

x =R

3

6

7 The sketch of the situation is as below:

By conservation of linear momentum,

3mvii−mvii = 3mv1f

(cos θ i+ sin θ j

)−mv2f j

2mvi i = 3mv1f cos θ i+ (3mv1f sin θ −mv2f ) j

Comparing components, we can obtain 2 equations:

2mvi = 3mv1f cos θ ⇒ vi =3

2v1f cos θ

mv2f = 3mv1f sin θ ⇒ v2f = 3v1f sin θ

Since the collision is elastic, the kinetic energy of the system is conserved.

1

2(3m) v2i +

1

2mv2i =

1

2(3m) v21f +

1

2mv22f

4v2i = 3v21f + v22f

4

(3

2v1f cos θ

)2

= 3v21f + (3v1f sin θ)2

9 cos2 θ = 3 + 9 sin2 θ

9(cos2 θ − sin2 θ

)= 3

cos 2θ =1

3θ = 35.3◦