Steel pellets, each with a mass of 0.60 g, fall vertically onto a horizontal plate at a rate of 100 pellets per minute.

Question 21

Steel pellets, each with a mass of 0.60 g, fall vertically onto a horizontal plate at a rate of 100 pellets per minute. They strike the plate with a velocity of 5.0 m s^-1 and rebound with a velocity of 4.0 m s^-1.

What is the average force exerted on the plate by the pellets?

A 0.0010 N                  B 0.0054 N                  C 0.0090 N                  D 0.54 N

Reference: Past Exam Paper – March 2018 Paper 12 Q10

Solution:

Answer: C.

Force is defined as the rate of change of momentum.

Force F = Δp / Δt

As the pellets strike the plate, their momentum changes. The velocity of the pellets changes direction as they rebound.

Consider one steel pellet,

Initial velocity = 5.0 m s^-1

Final velocity = – 4.0 m s^-1      (the direction changes)

Change in momentum Δp = mΔv = 0.6×10^-3 × (–4.0 – 5.0) = (–) 5.4×10^-3 Ns

This is the change in momentum of ONE steel pellet.

Rate of fall of pellets on the plate = 100 pellets per min

1 min - - > 100 pellets

60 s - - > 100 pellets

1 s -- > 100 / 60 = 1.667 pellets

In 1 second, an average of 1.667 pellets fall on the plate.

Force F = Δp / Δt        (force is the change in momentum per second)

Average force = rate of change of momentum of all pellets

Average force = 1.667 × 5.4×10^-3 = 0.0090 N

The Moon may be considered to be a uniform sphere of diameter 3.4 × 103 km and mass 7.4 × 1022 kg. The Moon has no atmosphere.

Question 6

(a) (i) A gravitational field may be represented by lines of gravitational force.

State what is meant by a line of gravitational force. [1]

(ii) By reference to lines of gravitational force near to the surface of the Earth, explain why the gravitational field strength g close to the Earth’s surface is approximately constant. [3]

(b) The Moon may be considered to be a uniform sphere of diameter 3.4 × 10³ km and mass 7.4 × 10²² kg. The Moon has no atmosphere.

During a collision of the Moon with a meteorite, a rock is thrown vertically up from the surface of the Moon with a speed of 2.8 km s^-1.

Assuming that the Moon is isolated in space, determine whether the rock will travel out into distant space or return to the Moon’s surface. [4]

 [Total: 8]

Reference: Past Exam Paper – June 2018 Paper 42 Q1

Solution:

(a) (i) A line of gravitational force represent the direction of the force on a (small test) mass placed in the field.

(ii) At the surface the lines of force are radial. Since the Earth has a large radius, a height above the surface is small (compared to the radius). So, the (radial) lines can be approximated as parallel lines close to the surface. Parallel lines imply that the field strength is constant.

(b)

{The threshold for escape is when the kinetic energy of the rock exceeds the potential energy required to reach infinity.

We need to compare the threshold value and the actual value to conclude whether the rock escape into space or not.}

{In this method we find the escape speed required by the rock to escape into space. If the speed of the rock (= 2.8×10³ m s^-1) is greater than the escape speed, then the rock would travel out into distant space.}

(change in) KE of rock = (change in) PE

½ mv² = GMm / R

{We need to convert values into SI units.

mv² = 2GMm / R }

(m)v² = (m)(2 × 6.67×10^-11 × 7.4×10²²) / (1.7×10³×10³)

v = 2.4 × 10³ m s^-1       

Since the speed of the rock is greater than the escape speed, the rock will travel out into distance space.