VCE Physics - Random Exam Revision 3

On 30 July 2020, the National Aeronautics and Space Administration (NASA) launched an Atlas rocket (Figure a) containing the Perseverance rover space capsule (Figure b) on a scientific mission to explore the geology and climate of Mars, and search for signs of ancient microbial life.

At lift-off from launch, the acceleration of the rocket was 7.20 m s⁻². The total mass of the rocket and capsule at launch was 531 tonnes.

Calculate the magnitude and the direction of the thrust force on the rocket at launch.

_{VCAA 2021}

• 3.82 MN, upwards

• 3.82 MN, downwards

• 9.03 MN, downwards

• 9.03 MN, upwards

Answer explanation

There are two forces acting on the rocket, the thrust upwards and the weight downwards.

       F_net = Thrust – Weight

531 × 10³ × 7.20=Thrust – 531 × 10³ × 9.8

      

Thrust = 531 × 10³(7.20 + 9.8)

Thrust = 9 027 000 N

       = 9.03 MN

The thrust is upwards.

 

9.03 MN, upwards

A car of mass 800 kg is towed along a straight road so that its velocity changes uniformly from 10 m s^-1 to 20 m s^-1 in a distance of 200 m. The frictional force is constant at 500 N.

Calculate the acceleration of the car.

_{VCAA 1977}

• 0.75 ms^-2

• 1.0 ms^-2

• 1.25 ms^-2

• 25 ms^-2

Answer explanation

v² = u² + 2ax

20² = 10² + 2 × a × 200

400 = 100 + 400a

300 = 400a

a = 0.75 ms^-2

Two blocks, each of mass m, are connected by means of a string which passes over a frictionless pulley. One is at rest on a frictionless table; the other is held at rest in the position shown.

What is the force of the string on Block X?

_{VCAA 1981}

• zero

• mg/2

• mg

• 2mg

Answer explanation

Block X is initially stationary and remains stationary until the hand supporting Block Y is removed. Therefore the net force on Block X is zero

Two blocks, each of mass m, are connected by means of a string which passes over a frictionless pulley. One is at rest on a frictionless table; the other is held at rest in the position shown.

Block Y is then released, and the blocks move. What will then be the net force on Y?

_{VCAA 1981}

• zero

• mg/2

• mg

• 2mg

Answer explanation

When Block Y is released, its weight accelerates it downward.

As the two blocks are connected by a string, they are going to both have the same acceleration when released.

The net force acting on Block Y is given by

mg – T = ma

The net force acting on Block X is given by

T = ma

Substituting for T gives

mg – ma = ma

                                   

mg = 2ma

                                   

ma = mg/2

A car is driving up a uniform slope with a trailer attached, as shown below. The slope is angled at 15⁰ to the horizontal. The trailer has a mass of 200 kg and the car has a mass of 750 kg.

Ignore all retarding friction forces down the slope.

Calculate the gravitational potential energy gained by the car and trailer when they have travelled 100 m along the slope.

_{VCAA 2021}

• 1.9 × 10⁵ J

• 2.4 × 10⁵ J

• 9.0 × 10⁵ J

• 9.3 × 10⁵ J

Answer explanation

∆GPE = mg∆h

GPE = (750 + 200) × 9.8 × 100sin15⁰

GPE = 240960.5

GPE = 2.4 × 10⁵ J

A 45 g golf ball, initially at rest, is hit by a golf club.

The contact time between the club and the ball is 0.50 ms.

The magnitude of the final velocity of the ball is 41 m s^–1.

Which one of the following is closest to the average force experienced by the golf ball?

_{VCAA 2021}

• 0.18 kN

• 0.37 kN

• 1.8 kN

• 3.7 kN

Answer explanation

F∆t = m∆v

F × 0.50 × 10^-3 = 45 × 10^-3 × 41

F = 3690 N

A ball of mass M strikes a stationary ball of mass m elastically, and head-on.

Which two of the following equations are correct?

_{VCAA 1979}

• U = v + V

• MU² = mv² + MV²

• M²U²

  = m²v² + M²+V²

• MU

  =(m + M)(v + V)

• MU = mv + MV

Answer explanation

Momentum is always conserved, and so is energy as the collision is elastic

A ball is projected from the ground at an angle of 30⁰ to the horizontal and at a speed of 40 m s^–1, as shown below. Ignore any air resistance.

Which one of the following graphs best shows the kinetic energy of the ball as a function of horizontal distance, d, from the launching point?

_{VCAA 2016}

• 
• 
• 
• 

Answer explanation

The speed of the ball is a minimum at the top of the flight. (It is not zero, as it still has a horizontal component). The total energy (KE + GPE) of the ball will remain constant. The Gravitational PE will have the shape of the motion, so the KE must be A

A ball is projected from the ground at an angle of 30⁰ to the horizontal and at a speed of 40 m s^–1, as shown below. Ignore any air resistance.

Calculate the distance, d, to the point where the ball hits the ground.

_{VCAA 2016}

• 69.3 m

• 139 m

• 103.9 m

• 209 m

Answer explanation

consider the vertical direction and find the time it takes to get to the top.

Use v = u – gt

0 = 40 × sin30 – 10 t

0 = 40 × 0.5 – 10 t

10 t = 20

t = 2 sec

it takes 4 secs for the total flight.

 

In the horizontal direction

Use d = v_horizontal × t

d = 40 cos30 × 4

d = 138.56

d = 139 m

An electron is accelerated from rest by a potential difference of V0. It emerges at a speed of 2.0 × 10⁷ m s^–1 into a magnetic field, B, of strength 2.5 × 10^–3 T and follows a circular arc, as shown.

Calculate the radius of the path travelled by the electron.

_{VCAA 2021 NHT}

• 1.2x10^-45 m

• 3.8 x 10^-2 m

• 4.6 x 10^-2 m

• 4.6 x 10¹⁶ m

Answer explanation

F = mv²/r and F = Bqv

Bq = mv/r

r = mv/Bq

r = 9.1x10^-31 x 2x10⁷ /

(2.5x10^-3 x 1.6 x 10^-19)

      

r = 0.0455

r = 0.046 m