Questions on motion

Question

If a question states 'from rest' what information about the equations is being given?

Initial speed u = 0.

Question

If a question states 'freely under gravity', what information about the equations is being given?

Acceleration = 9.8 ms-2 or -9.8 ms-2 due to gravity.

Remember to have all your vectors in one direction as positive and make all the vectors in the opposite direction negative.

Question

An object lifts off vertically and takes a time of 4s to reach the highest point. What time will it take to fall back to Earth?

4 seconds – assuming no rocket force or air resistance. The acceleration due to gravity is the same in both directions, slowing down on the way up, speeding up on the way back down.

Question

A ball is projected vertically upwards with an initial velocity of 24.5 ms-1.

The effects of air resistance may be neglected.

Calculate the time taken for the ball to reach its maximum height.

Vector diagram where upwards is positive direction for vectors. U equals 24.5 metres per second upwards. V equals zero. A equals 9.8 metres per second per second downwards. T and s are both unknown.

Let unknown vector quantities go in the positive direction for vectors.

Vertical motion is a constant acceleration of -9.8 ms-2 towards the centre of the Earth.

u = 24.5m{s^{ - 1}} (upwards)

v = 0 (stationary at top)

a =  - 9.8m{s^{ - 2}} (acceleration is negative as it is in downwards direction)

t = ? This is the unknown we are trying to find.

s = ? As displacement is unknown we need to use an equation that does not include displacement.

v = u + at

0 = 24.5 + ( - 9.8)t

t = 2.5s

Question

Now calculate the maximum height reached by the ball to the nearest metre.

u = 24.5m{s^{ - 1}}

v = 0

a =  - 9.8m{s^{ - 2}}

t = 2.5s

s = ? This is the unknown we need to find.

s = ut + \frac{1}{2}a{t^2}

s = (24.5 \times 2.5) + \frac{1}{2}( - 9.8) \times {2.5^2}

s = 61.25 - 30.625

s = 30.625

Maximum height reached is 31 m.

Question

A nail is fired from a nail gun into a fixed block of wood. The nail has a speed of 380 ms-1 just as it enters the wood.

Vector diagram where left to right is the positive direction for vectors. U equals 380 metres per second left to right. V equals zero. S equals 60 millimetres. A and t are both unknown.

The nail comes to rest after penetrating 60 mm into the wood.

Find the time taken for the nail to come to rest. Assume that the frictional force on the nail is constant as it penetrates the wood.

u = 380m{s^{ - 1}}

v = 0

a = ?

t = ?

s = 60 \times {10^{ - 3}}m

s= \frac{{u + v}}{2}t

t=\frac{2s}{u+v}

t = \frac{{2\times 0.06}}{{380}} = 3.16 \times {10^{- 4}}s

The time taken for the nail to come to rest is 3.16 x 10-4s.