1) How is stopping distance calculated?

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Presentation transcript:

1) How is stopping distance calculated? Answer Thinking distance + braking distance Challenge What is the definition of thinking distance? Challenge Answer The distance travelled in the time it takes the driver to react to a situation.

2) Name 3 things that affect braking distance? Answer Road condition, tyre condition, brake condition, weather condition, speed, gradient of road etc Challenge How does speed affect braking distance? Challenge Answer The faster the car is travelling, the further it will travel once the brakes have been applied.

3) Name 3 things that affect thinking distance? Answer Distractions, drugs, alcohol, speed Challenge How does speed affect thinking distance? Challenge Answer The faster the car is travelling, the further it will travel in the time it takes the driver to react.

4) A bus takes 10m to stop after the driver applies the brakes 4) A bus takes 10m to stop after the driver applies the brakes. The brakes exert a force of 1000N. Calculate how much work is done by this force. Answer E = F x d 1000 x 10 = 10,000J Challenge What is the energy transferred during this process? Challenge Answer 10,000J because work done equals energy transferred

5) If the bus in the previous Q takes 4s to stop, calculate the power. Answer P = E/t 10,000/4 = 2,500W

If the car falls back down to the floor, what is the gain in KE? 6) A car of mass 1000kg is raised 25m above the ground. Calculate the gain in GPE in this scenario. Answer GPE = m x g x h (where g = 10 on Earth) 1000 x 10 x 25 = 250,000J Challenge If the car falls back down to the floor, what is the gain in KE? Challenge Answer 250,000J because the loss in GPE is equal to the gain in KE

What is the work done on the car? 7) The same car from the previous Q is raised into the air, gaining 50,000J of GPE. How high has it been raised? Answer h = GPE/(m x g) 50,000/(1000 x 10) = 50,000/10,000 = 5m Challenge What is the work done on the car? Challenge Answer 50,000J because work done equals energy transferred.

8) If the car from the previous Q falls back to the floor with of loss of 50,000J of GPE, how much KE is gained? Answer 50,000J Challenge Calculate the velocity of the car Challenge Answer KE = ½ mv2 v = 2KE/m = (2 x 50,000)/1000 = 100 = 10m/s

Momentum before collision = momentum after collision 9) A mother pushes a pushchair of 10kg at a velocity of 2m/s. Calculate the momentum of the pushchair. Answer p = m x v 10 x 2 = 20kgm/s Challenge If the same push chair crashes into another pushchair of the same mass, travelling in the opposite direction with a speed of 1m/s, calculate the velocity of the pushchairs after the collision. Challenge Answer Momentum before collision = momentum after collision Total Momentum before = (10 x 2) + (10 x -1) = 20 – 10 = 10kgm/s Combined mass is 20kg Velocity after (v = p/m) = 10/20 = 0.5 m/s

10) A box is dropped from a height onto the floor 10) A box is dropped from a height onto the floor. How can we reduce the force of impact on the box as it hits the floor? Explain your answer. Answer Wrap the box in a cushioned material or put bubblewrap inside the box around the object. Air cushions/cushioned material increases the duration of the impact, therefore decreasing the force of impact. It follows the principle of the following equation: F = m ( v – u ) t

11) Exam Question: A child on a swing is raised up in the air and released. The kinetic energy of the child on the swing varies over the next few swings. The total energy of the swing also decreases over time. Explain the energy changes during this scenario. Answer KE varies during swing KE max at bottom of swing, minimum at the top of its swing GPE varies during swing GPE max at top of swing, minimum at the bottom of its swing There is an interchange of KE and GPE throughout the swing, keeping the energy in one swing constan Over a number of swings, the total energy in the swing decreases because energy is dissipated/lost to the environment. This is due to friction and air resistance slowing the swinging child down therefore the height the child reaches with each swing decreases with each successive swing.