Further Mechanics - collisions, momentum and impulse Flashcards
Derivation of the impulse equation from Newton’s 2nd Law
F=ma , ∆p=F∆t
F = m a (N2L) F = m (∆v / ∆t) (a = ∆v/∆t) F = (m∆v) / ∆t (m x ∆v / ∆t = m∆v / ∆t) F = ∆p / ∆t (m x ∆v = ∆p) F x ∆t = ∆p (Rearrange) ∆p = F∆t (Impulse)
Unit for ∆p
Ns
Unit for p
kg ms^-1
Elastic Collision are when …
… collisions have no loss in kinetic energy (KE is conserved within the system).
Inelastic collisions are when …
… collisions result in a change in kinetic energy within the system. This energy is then transferred to a different store (Internal/thermal etc).
Derivation of Ek = p^2 / 2m from the kinetic energy and momentum equation
Ek = 1/2 m v^2, p = m v
p^2 = m^2 v^2 ( (p=mv)^2 )
p^2 / m^2 = v^2 (Rearrange)
Ek = 1/2 m v^2 Ek = 1/2 m (p^2 / m^2) (v^2 = p^2 / m^2) Ek = ( 1 x m x p^2 ) / 2 m^2 Ek = p^2 / 2m ( m/2m^2 = 1/ 2m)
What can be calculated from the area under a force - time graph?
Impulse
How do you find the angle, when you have arc length and radius?
θ = arc / radius
A nucleus decays by releasing an alpha particle. The speed of the recoiling nucleus is small compared to the speed of the alpha particle. This is due to…
… the recoiling nucleus has a much larger mass than the alpha particle.
The area under a force-time graph represents?
Change in momentum
Principle of conservation of momentum states that…
… in a closed system, the net momentum before an action is equal to the net momentum after an action.
A 24.2 g bullet enters a 2.81 kg watermelon and embeds itself in the melon. The melon is immediately set into motion with a speed of 3.82 ms-1. The bullet remains lodged inside the melon. What was the entry speed of the bullet?
v of bullet = 447 ms^-1
Newton’s first law states that…
… an object in motion will stay in motion and a stationary object will remain stationary, unless acted on by an external resultant force.
Newton’s second law states that…
… the acceleration of an object is directly proportional to the net force applied and inversely proportional to its own mass.
N2L equation is…
… net F = ma
Newton’s third law states that…
… for every action, in an interaction pair, there is always an equal and opposite reaction.
Ball A has a mass of 10 kg and moves with an initial velocity of 3.0 ms^-1 due east. Ball B has a mass of 5.0 kg and is stationary.
Ball A collides with ball B and moves away at an angle of 30 degrees to the horizontal, with a new velocity of 2.0 ms^-1.
What is the magnitude and direction of the velocity that ball B will move away with?
v of mass B = 3.23 ms^-1
at 38.3 degrees below horizontal
In an elastic collision what two things are conserved?
KE and Momentum
A 1.2 kg ball, A, is moving with a velocity of 4.4 ms^-1 due west. It collides with a stationary ball, B, also with a mass of 1.2 kg. After the collision ball A moves off at 38 degrees south of west while ball B moves off at 53 degrees north of west.
Find the velocities of both balls after the collision.
Final v of B = 10.5 ms^-1
Final v of A = -13.6 ms^-1
In an elastic collision what two things are conserved?
KE and Momentum
