Questions on motion
Question
If a question states 'from rest' what information about the equations is being given?
The speed of a body before accelerating. Eg: A body has an initial speed of 5 m s-1. 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 鈥 assuming no A rocket is usually a chemically fueled vehicle that exerts a large force while burning a small volume of fuel. Generally used to launch objects high into the atmosphere or into space. Eg: the rocket launch allowed the satellite to be put into orbit. force or A force of friction produced when an object moves through the air.. 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.
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.
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.