A cannon fires a shell vertically upwards. The shell leaves the cannon with a speed of 80 m s-1 and a kinetic energy of 480 J.

Question 33

(a) State what is meant by kinetic energy. [1]

(b) A cannon fires a shell vertically upwards. The shell leaves the cannon with a speed of 80 m s^-1 and a kinetic energy of 480 J. The shell then rises to a maximum height of 210 m. The effect of air resistance is significant.

(i) Show that the mass of the shell is 0.15 kg. [2]

(ii) For the movement of the shell from the cannon to its maximum height, calculate

1. the gain in gravitational potential energy, [2]

2. the work done against air resistance. [1]                     

(iii) Determine the average force due to the air resistance acting on the shell as it moves

from the cannon to its maximum height. [2]

(iv) The shell leaves the cannon at time t = 0 and reaches maximum height at time t = T.

On Fig. 2.1, sketch the variation with time t of the velocity v of the shell from time t = 0 to

time t = T. Numerical values of v and t are not required.

Fig. 2.1

[2]

(v) The force due to the air resistance is a vector quantity.

Compare the force due to the air resistance acting on the shell as it rises with the force

due to the air resistance as it falls. [2]

[Total: 12]

Reference: Past Exam Paper – November 2018 Paper 23 Q2

Solution:

(a) Kinetic energy is the energy (of a mass/body/object) due to motion/speed/velocity

(b)

(i)

E = ½ mv²                  

480 = ½ × m × 80²      

m = 0.15 kg

(ii)

1.

E = mgh          or ΔE = mgΔh

E = 0.15 × 9.81 × 210

E = 310 J

2.

{KE of shell at cannon = GPE of cannon at top + Work done against air resistance

480 = 310 + Work done}

work done = 480 – 310

work done = 170 J

(iii)

{Work done is equal to the product of air resistance and the vertical distance travelled by the shell.}

work done = Fs

force = 170 / 210                                           

force = 0.81 N

(iv)

curved line from positive value on v-axis to (T, 0)

magnitude of gradient decreases

{Initially, the speed of the shell is 80 m s^-1. As the shell moves up, its speed decreases as the force of gravity is acting downwards. So, the value of velocity decreases with time. At the maximum height (at time T), the velocity is zero.

The acceleration due to gravity is downwards, opposite to the upward motion. The acceleration is given by the gradient of the velocity-time graph, so the gradient is negative. The shell is decelerating.

However, the deceleration is not constant as the force of gravity is not the only force opposite the motion. Air resistance is also opposing the motion. Air resistance depends on the speed of motion. Since the speed is decreasing with time, the air resistance also decreases with time. So, the decrease in speed becomes less as the shell gains height (since its speed is decreasing). The gradient of the graph is decreasing with time.}

(v)

As the shell rises the force decreases and as the shell falls the force increases

As the shell rises the force is downward and as the shell falls the force is upward

{Force is a vector and has both a magnitude and a direction. So in this comparison, we need to describe the magnitude and direction of the force.

Air resistance always opposes the motion.

As the shell rises, the force is downward (opposite to motion). The speed of the shell decreases with time, so the air resistance also decreases as the shell rises.

As the shell falls, the force is upward (opposite to motion). The speed of the shell increases with time, so the air resistance also increases as the shell falls.}