VCE Physics - Random Exam Revision 2

Students set up an experiment as shown. M₁, of mass 4.0 kg, is connected by a light string (assume it has no mass) to a hanging mass, M₂, of 1.0 kg. The system is initially at rest. Ignore mass of string and friction. The masses are then released from rest.

Calculate the acceleration of M₁.

VCAA 2015

• 0.51 ms^-1

• 1.96 ms^-1

• 7.84 ms^-1

• 9.80 ms^-1

Answer explanation

The gravitational force on M₂ accelerates the entire system.

M₂g=(M₁ + M₂)a

1 x 9.8 = (4 + 1) a

a = 9.8 / 5

a = 1.96 ms^-2

Students set up an experiment as shown. M₁, of mass 4.0 kg, is connected by a light string (assume it has no mass) to a hanging mass, M₂, of 1.0 kg. The system is initially at rest. Ignore mass of string and friction. The masses are then released from rest.

If the acceleration of M₁ is 1.96 ms^-1, calculate the tension in the string.

VCAA 2015

• 0.51 N

• 1.96 N

• 7.84 N

• 9.80 N

Answer explanation

T= M₁a

T = 4 x 1.96

T = 7.84N

alternative method:

M₂g - T=M₂a

1 x 9.8 - T= 1 x 1.96

9.8 - T = 1.96

T = 9.8-1.96

T = 7.84 N

A model car of mass 2.0 kg is on a track that is part of a vertical circle of radius 4.0 m, as shown. At the lowest point, L, the car is moving at 6.0 m s–1. Ignore friction.

Which of the following shows the resultant force acting on the car at the lowest point ?

_{VCAA 2015}

• 
• 
• 
• 

Answer explanation

Uniform circular motion, the net force acts radially inwards.

In a cricket game, a batsman hits a ball at an angle of 30° to the horizontal at a speed of 20 m s^–1. The playing area and the area around it are perfectly flat, and the ball is assumed to be at ground level when the batsman hits it. Ignore air resistance.

Calculate the maximum height reached by the ball.

_{VCAA 2014}

• 0.51 m

• 5.1 m

• 10.2 m

• 20.5 m

Answer explanation

The vertical component of the initial velocity:

20sin30 = 10 m s ^–1

vertical motion:

v ² = u ² + 2as

0² = 10² + 2 x -9.8 x s

0 = 100 - 19.6 s

s = 100/19.6 = 5.1 m

A distant star has a planet orbiting it. The period of the planet’s circular orbit is 1200 hours. The radius of the planet’s orbit is measured to be 7.0 × 10¹⁰ m.

Calculate the mass of the star.

_{VCAA 2014}

• 1.4 x 10³⁸

• 4.7 x 10³⁷

• 3.5 x 10³⁰

• 1.1 x 10³¹

Answer explanation

GMm/R² = 4π²Rm/T²

Transposing gives:

M = 4π²R ³ /GT²

M = 4π²R ³ /GT²

T = 1200 hours = 1200 x 3600 s

T = 4.32 x 10⁶ s

R = 7.0 x 10¹⁰ m

M = 4π²(7.0 x 10¹⁰)³ /G(4.32 x 10⁶)²

M = 1.09 x 10³¹

A spring is resting against a wall. The spring is compressed by a distance of 8.0 cm from its uncompressed length. Jemima holds a block of mass 1.2 kg stationary against the compressed spring as shown.

Jemima releases the block and it slides to the right on a frictionless surface. It leaves the spring with a kinetic energy of 5.4 J and slides at constant speed

The speed of the block as it leaves the spring is closest to:

_{VCAA 2012}

• 2 ms^-1

• 3 ms^-1

• 4.6 ms^-1

• 9 ms^-1

Answer explanation

KE = 0.5mv²

5.5=0.5 x 1.2 x v²

v² = 9.1

v = 3 ms^-1

A spring is resting against a wall. The spring is compressed by a distance of 8.0 cm from its uncompressed length. Jemima holds a block of mass 1.2 kg stationary against the compressed spring as shown.

Jemima releases the block and it slides to the right on a frictionless surface. It leaves the spring with a kinetic energy of 5.4 J and slides at constant speed

Calculate the work done by the spring on the block.

_{VCAA 2012}

• 0 J

• 4.5 J

• 5.4 J

• 11.8 J

Answer explanation

The work done by the spring was equal to the kinetic energy provided to the block, 5.4 J.

A spring is resting against a wall. The spring is compressed by a distance of 8.0 cm from its uncompressed length. Jemima holds a block of mass 1.2 kg stationary against the compressed spring as shown.

Jemima releases the block and it slides to the right on a frictionless surface. It leaves the spring with a kinetic energy of 5.4 J and slides at constant speed

Calculate the spring constant, k, of the spring. Assume that the spring obeys Hooke’s law

_{VCAA 2012}

• 1.7 × 10³ N m^–1

• 9.8 N m^–1

• 1.3 × 10² N m^–1

• 67.5 N m^-1

Answer explanation

The kinetic energy provided was equal to the energy stored in the spring.

5.4 = ½ k x ² , x=0.08m

k = 1.7 × 10³ N m^–1 .

A spring is resting against a wall. The spring is compressed by a distance of 8.0 cm from its uncompressed length. Jemima holds a block of mass 1.2 kg stationary against the compressed spring as shown.

Jemima releases the block and it slides to the right on a frictionless surface. It leaves the spring with a kinetic energy of 5.4 J and slides at constant speed of 3 ms^-1

Calculate the magnitude of the total impulse that the spring gives to the 1.2 kg block by the time it leaves the spring.

_{VCAA 2012}

• 1.7 × 10³ Ns

• 9.8 Ns

• 4.5 Ns

• 3.6 Ns

Answer explanation

The impulse given to the block by the spring was equal to the change in momentum of the block.

Impulse = m Δv = 3.6 N s.

A metal ring is to be held stationary by three forces that are all pulling on the ring. All the forces are greater than zero, but their magnitudes are not given. Possible directions of the forces on the ring are shown in the arrangements. Only one of these arrangements can hold the ring stationary. Identify which one

_{VCAA 2012}

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Answer explanation

Option B was correct because it was the only arrangement that could produce a net force of zero.