Numerical Problems - Physics Solutions (Part 3)

Interactive Animation: Collision Programmed

Use this interactive simulation directly inside this chapter page.

Problem 3.1: Block on a smooth horizontal surface

A 10 kg block is placed on a smooth horizontal surface. A horizontal force of 5 N is applied to the block. Find:

(a) the acceleration produced in the block.

(b) the velocity of block after 5 seconds.

GIVEN

Mass: \( m = 10 \, \text{kg} \)

Force: \( F = 5 \, \text{N} \)

Time: \( t = 5 \, \text{s} \)

REQUIRED

Acceleration \( a \) = ?

Final velocity \( v_f \) = ?

SOLUTION

(a) Acceleration

Using Newton's second law:

\( F = ma \)

Putting values:

\( a = \frac{F}{m} = \frac{5 \, \text{N}}{10 \, \text{kg}} = 0.5 \, \text{m/s}^2 \)

(b) Velocity after 5 seconds

Using first equation of motion:

\( v_f = v_i + a t \)

Putting values:

\( v_f = 0 + (0.5 \, \text{m/s}^2)(5 \, \text{s}) = 2.5 \, \text{m/s} \)

[Diagram for 3.1: Block on smooth surface with applied force]

Answer: (a) \( a = 0.5 \, \text{m/s}^2 \), (b) \( v = 2.5 \, \text{m/s} \)

Problem 3.2: Weight of a person

The mass of a person is 80 kg. What will be his weight on the Earth? What will be his weight on the Moon? The value of acceleration due to gravity of Moon is 1.6 m s\(^{-2}\).

GIVEN

Mass: \( m = 80 \, \text{kg} \)

Earth's gravity: \( g_e = 10 \, \text{m/s}^2 \)

Moon's gravity: \( g_m = 1.6 \, \text{m/s}^2 \)

REQUIRED

Weight on Earth \( w_e \) = ?

Weight on Moon \( w_m \) = ?

SOLUTION

(a) Weight on Earth

The weight of a body is:

Using: \( w = mg \)

Putting values:

\( w_e = (80 \, \text{kg})(10 \, \text{m/s}^2) = 800 \, \text{N} \)

(b) Weight on Moon

Putting values:

\( w_m = (80 \, \text{kg})(1.6 \, \text{m/s}^2) = 128 \, \text{N} \)

[Diagram for 3.2: Person on Earth and Moon showing different weights]

Answer: Weight on Earth = 800 N, Weight on Moon = 128 N

Problem 3.3: Force required to increase the velocity

What force is required to increase the velocity of 800 kg car from 10 m s\(^{-1}\) to 30 m s\(^{-1}\) in 10 seconds?

GIVEN

Mass: \( m = 800 \, \text{kg} \)

Initial velocity: \( v_i = 10 \, \text{m/s} \)

Final velocity: \( v_f = 30 \, \text{m/s} \)

Time: \( t = 10 \, \text{s} \)

REQUIRED

Force \( F \) = ?

SOLUTION

First, find acceleration:

\( a = \frac{v_f - v_i}{t} = \frac{30 - 10}{10} = 2 \, \text{m/s}^2 \)

Now, find force:

\( F = ma = (800 \, \text{kg})(2 \, \text{m/s}^2) = 1600 \, \text{N} \)

[Diagram for 3.3: Car accelerating with applied force]

Answer: \( F = 1600 \, \text{N} \)

Problem 3.4: Bullet fired from a gun

A 5 g bullet is fired by a gun. The bullet moves with a velocity of 300 m s\(^{-1}\). If the mass of the gun is 10 kg, find the recoil speed of the gun.

GIVEN

Mass of bullet: \( m_b = 5 \, \text{g} = 0.005 \, \text{kg} \)

Velocity of bullet: \( v_b = 300 \, \text{m/s} \)

Mass of gun: \( m_g = 10 \, \text{kg} \)

REQUIRED

Recoil velocity: \( v_g = ? \)

SOLUTION

Using conservation of momentum:

\( m_b \, v_b + m_g \, v_g = 0 \)

or:

\( v_g = -\frac{m_b \, v_b}{m_g} \)

putting values:

\( v_g = -\frac{0.005 \, kg \times 300 \, m/s}{10 \, kg} \) = - \( \frac{1.5 \, kg \, m/s}{10 \, kg} = -0.15 \, \text{m/s} \)

[Diagram for 3.4: Gun firing bullet with recoil]

Answer: \( v_g = -0.15 \, \text{m/s} \)

Problem 3.5: Astronaut & a wrench

An astronaut weighs 70 kg. He throws a wrench of mass 300 g at a speed of 3.5 m s\(^{-1}\). Determine:

(a) the speed of astronaut as he recoils away from the wrench.

(b) the distance covered by the astronaut in 30 minutes.

GIVEN

Mass of astronaut: \( m_a = 70 \, \text{kg} \)

Mass of wrench: \( m_w = 300 \, \text{g} = 0.3 \, \text{kg} \)

Velocity of wrench: \( v_w = 3.5 \, \text{m/s} \)

Time: \( t = 30 \, \text{min} = 1800 \, \text{s} \)

REQUIRED

Recoil speed of astronaut: \( v_a = ? \)

Distance covered: \( S = ? \)

SOLUTION

(a) Recoil speed of astronaut

Using conservation of momentum:

\( m_a v_a + m_w v_w = 0 \)

or:

\( v_a = -\frac{m_w \times v_w}{m_a} \)

putting values:

\( v_a = -\frac{0.3 \, kg \times 3.5 \, m/s}{70 \, kg} \)

therefore:

\( v_a = -\frac{1.05 \, kg \, m/s }{70 \, kg} = -0.015 \, \text{m/s} \)

hence:

\( v_a = -1.5 \times 10^{-2} \, \text{m/s} \)

(b) Distance covered in 30 minutes

\( S = v_a \times t = (0.015 \, \text{m/s})(1800 \, \text{s}) = 27 \, \text{m} \)

[Diagram for 3.5: Astronaut throwing wrench in space]

Answer: (a) \( v_a = -1.5 \times 10^{-2} \, \text{m/s} \), (b) \( S = 27 \, \text{m} \)

Problem 3.6: Bogie of a goods train

A 6.5 × 10\(^3\) kg bogie of a goods train is moving with a velocity of 0.8 m s\(^{-1}\). Another bogie of mass 9.2×10\(^3\) kg coming from behind with a velocity of 1.2 m s\(^{-1}\) collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.

GIVEN

Mass of first bogie: \( m_1 = 6.5 \times 10^3 \, \text{kg} \)

Velocity of first bogie: \( v_1 = 0.8 \, \text{m/s} \)

Mass of second bogie: \( m_2 = 9.2 \times 10^3 \, \text{kg} \)

Velocity of second bogie: \( v_2 = 1.2 \, \text{m/s} \)

REQUIRED

Combined velocity of bogie: \( v = ? \)

SOLUTION

Using conservation of momentum::

\( m_1 v_1 + m_2 v_2 = (m_1 + m_2)v \)

putting values:

\( (6500 \, kg) \, (0.8 \, m/s ) + (9200 \, kg ) \, (1.2 \, m/s ) = (6500 \, kg + 9200 \, kg ) \, v \)

or:

\( 5200 \, kg \, m/s + 11040 \, kg \, m/s = 15700 \, kg \, \times v \)

Hence:

\( v = \frac{16240 \, kg \, m/s }{15700 \, kg } = 1.034 \, \text{m/s} \)

[Diagram for 3.6: Two train bogies coupling together]

Answer: \( v = 1.03 \, \text{m/s} \)

Problem 3.7: A cyclist

A cyclist weighing 55 kg rides a bicycle of mass 5 kg. He starts from rest and applies a force of 90 N for 8 seconds. Then he continues at a constant speed for another 8 seconds. Calculate the total distance travelled by the cyclist.

GIVEN

Mass of cyclist: \( 55 \, \text{kg} \)

Mass of bicycle: \( 5 \, \text{kg} \)

Total mass: \( m = 60 \, \text{kg} \)

Force: \( F = 90 \, \text{N} \)

Time for acceleration: \( t_1 = 8 \, \text{s} \)

Time at constant speed: \( t_2 = 8 \, \text{s} \)

REQUIRED

Total distance covered: \( S = ? \)

SOLUTION

First phase (accelerating): Distance Covered

First phase: Acceleration: To determine distance covered in the first phase we will find the acceleration first:

\( a = \frac{F}{m} = \frac{90}{60} = 1.5 \, \text{m/s}^2 \)

We will now determine distance covered in the first phase by using second equation of motion:

\( S_1 = v_i t + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1.5)(8)^2 = 48 \, \text{m} \)

Second phase (Constant speed): Distance Covered

To determine the distance covered in the second phase, we will first determine the final velocity:

\( v_f = v_i + at = 0 + (1.5)(8) = 12 \, \text{m/s} \)

Distance in second phase:

\( S_2 = v_f t_2 = (12)(8) = 96 \, \text{m} \)

Total distance

\( S_{total} = S_1 + S_2 = 48 + 96 = 144 \, \text{m} \)

[Diagram for 3.7: Cyclist motion with acceleration and constant speed phases]

Answer: \( S_{total} = 144 \, \text{m} \)

Problem 3.8: Ball rebounding

A ball of mass 0.4 kg is dropped on the floor from a height of 1.8 m. The ball rebounds straight upward to a height of 0.8 m. What is the magnitude and direction of the impulse applied to the ball by the floor?

GIVEN

Mass: \( m = 0.4 \, \text{kg} \)

Initial height: \( h_1 = 1.8 \, \text{m} \)

Rebound height: \( h_2 = 0.8 \, \text{m} \)

Acceleration due to gravity: \( g = 10 \, \text{m/s}^2 \)

REQUIRED

Impulse: \( J = ? \)

SOLUTION

Using the 3rd equation for ball falling down (free fall):

\( 2gS = v_f^2 - v_i^2 \)

First, find velocity just before hitting the floor, here \( v_i = 0 \), \( S = h_1 \) and \( v_f = v_1 \):

\( 2gS = v_1^2 \)

Taking square root on both sides:

\( v_1 = \sqrt{2gh_1} = \sqrt{2 \times 10 \times 1.8} = \sqrt{36} = 6 \, \text{m/s} \) (downward)

First, find velocity just after rebound, again we will use the third equation of motion but taking \( v_f = 0 \), \( g = -g \) , \( S = -h_2 \) and \( v_i = - v_2 \):

\( - 2gS = - v_2^2 \)

Taking square root on both sides:

\( v_2 = \sqrt{2gh_2} = \sqrt{2 \times 10 \times 0.8} = \sqrt{16} = 4 \, \text{m/s} \) (upward)

Taking upward as positive:

Initial momentum: \( p_i = m(-v_1) = 0.4 \times (-6) = -2.4 \, \text{N s} \)

Final momentum: \( p_f = m(v_2) = 0.4 \times 4 = 1.6 \, \text{N s} \)

Impulse = Change in momentum:

\( J = p_f - p_i = 1.6 - (-2.4) = 4 \, \text{N s} \)

[Diagram for 3.8: Ball dropping and rebounding with impulse]

Answer: Magnitude = 4 N s, Direction = upward

Problem 3.9: Collision of two balls

Two balls of masses 0.2 kg and 0.4 kg are moving towards each other with velocities \( 20 \, \text{m} \, \text{s}^{-1} \) and \( 5 \, \text{m} \, \text{s}^{-1} \) respectively. After collision, the velocity of 0.2 kg ball becomes \( 6 \, \text{m} \, \text{s}^{-1} \). What will be the velocity of 0.4 kg ball?

GIVEN

Mass of ball 1: \( m_1 = 0.2 \, \text{kg} \)

Initial velocity of ball 1: \( u_1 = 20 \, \text{m/s} \)

Final velocity of ball 1: \( v_1 = 6 \, \text{m/s} \)

Mass of ball 2: \( m_2 = 0.4 \, \text{kg} \)

Initial velocity of ball 2: \( u_2 = -5 \, \text{m/s} \)

REQUIRED

Final velocity of ball 2: \( v_2 = ? \)

SOLUTION

Note: Taking direction of \( m_1 \) as positive, \( u_2 = -5 \, \text{m/s} \)

Using conservation of momentum: \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \)

\( (0.2)(20) + (0.4)(-5) = (0.2)(6) + (0.4)v_2 \)

\( 4 - 2 = 1.2 + 0.4v_2 \)

\( 2 = 1.2 + 0.4v_2 \)

\( 0.4v_2 = 2 - 1.2 = 0.8 \)

\( v_2 = \frac{0.8}{0.4} = 2 \, \text{m/s} \)

[Diagram for 3.9: Two balls colliding with given velocities]

Answer: \( v_2 = 2 \, \text{m/s} \)