Topic 6a - Impulse And Momentum Model Answers Flashcards
(12 cards)
Define impulse
Change in momentum
Calculate the force exerted on the ball by the wall (see model answers pack of images)
Force = change in momentum/time
= -6-9 / (5 x 10 ^-3)
= 15/ (5 x 10 ^-3)
= 3000N
State the principle of conservation of momentum
The total momentum before and after a collision provided no external forces act
Applying newton’s second and third laws of motion to the collision between two pucks leads to the conclusion that momentum is conserved. Justify this statement
Applying Newton’s third law, when one puck A exerts a force on the other puck B, puck B exerts and equal and opposite force on puck A, we can write this as FA = -Fb
Due to Newton’s second law, F = delta p/ delta t so delta pa/delta ta = - deltapb/delta tb
The forces act for the same time (delta ta = delta tb)
So delta pa = -delta pb
Therefore delta Pa + delta pb = 0 so the total change in momentum is 0, momentum is conserved
Define inelastic collision
A collision where the total kinetic ENEGRY after the collision is less than the total kinetic energy before the collision
Describe some examples of inelastic collisions
When objects stick together, this must be an inelastic collision
If an object changes shape, work will have been done to change the shape of the object, so the kinetic ENEGRY will be transferred to the internal energy store of the material and this will be an inelastic collision
Define elastic collision
A collision where the total kinetic ENEGRY before and after a collision is the same
A piece of plasticine is fired from a spring gun, it travels through the air and hits a wooden pendulum bob and sticks too it the bob and clay move yup to a maximum height. Describe how the principle of energy conservation and the principle of momentum conservation apply to this situation
- The spring stores elastic potential ENEGRY which is transferred to the kinetic energy of the clay
- During the collision, the clay deforms and sticks to the bob so this is an inelastic collision and kinetic ENEGRY is not conserved - work is done to change the shape of the clay, increasing its internal energy (thermal energy)
- The bob and clay have kinetic energy as they move together, and this kinetic energy is transferred to GPE, until at maximum height all the KE has become GPE
- Momentum conservation applies during the collision
- The momentum of the clay before collision is equal to the momentum of the clay+ bob after the collision. The speed of the combined clay+ bob must be less than that of the clay alone
Two identical spheres of mass m are both travelling with a speed v towards each other
The spheres collide head on
What is the total momentum
Total momentum = 0
Whilst moving, empty fuel tanks can be ejected by means of an explosion, this has the fact of increasing the speed of the probe. Discuss whether conservation of momentum and conservation of energy apply in this situations nd why the speed of the probe increases
Conservation of momentum applies
The probe and tank experience equal and opposite changes in momentum
They change their speed in opposite direction
Total energy is conserved
But the kinetic ENEGRY of teh system increases
This means change Al ENEGRY must have been converted into kinetic energy
A moving ball A collides wuth an identical stationary ball B on ice. It makes a direct hit and the collusion is elastic. Just before the collision ball A has a vecloty v. After the collision ball B has a vecloty v. Discuss how the relevant conservation laws apply to this collision
The collision takes place on ice so there is minimal friction - external forces are negligible
So momentum is conserved
The momentum of ball A before the collision equals the momentum of ball A+B after the collision
As B has the same momentum as A did before ball A must be at rest after the collision
All of the kinetic ENEGRY of ball A must be transferred to ball B
As kinetic energy is conserved in an elastic collision
Derive the expression for KE in terms of momentum
P = mv
V = p/m
KE = 0.5mv^2
KE = 0.5 m (p/m)^2
KE = p^2/2m
