## Friday, 11 June 2010

### Mechanics III

The affect of collision depends on the mass and velocity of the object, so a physical quantities is created as momentum, which is the product of mass and velocity. Mathematically p=mv, which is a vector with unit kg ms-1.
By F=ma=m(Δv/t)=( Δmv)/t, we can conclude that F = change in momentum / time of action. In other works, the net force acting on the object is equal to the rate of change of momentum of it.
Impulse is also another physical quantities that Ft=Δ(mv), which is a vector with unit Ns.
Object deforms during collision, and the force that make it deforms is called impact force while that t is called the time of impact. The longer time of impact results in smaller impact force.
We can draw a force-time graph (F-t)graph to represent the collision, while ∫F dt=Δmv which is the impulse delivered to the object, while it is also the change of momentum of the object.
The law of conservation of momentum states that the total of momentum of a system is conserved if no net external force is acting on it. Mathematically mAuA+mBuB=mAvA+mBvB.
The collision can be classified as following:
1) Elastic collision: Total K.E. conserved. i.e., mAuA2/2+mBuB2/2=mAvA2/2+mBvB2/2.
2) Inelastic collision: Some K.E. lost : i.e., mAuA2/2+mBuB2/2>mAvA2/2+mBvB2/2
3) Perfectly inelastic collision: vA=vB, then K.E. lost is maximized.
Note that total K.E. won’t increase during collision if there’s no net external force.
Explosion: In this case chemical energy was transferred into K.E. The momentum can be given by mAvA=-mBvB.
Apparent loss of momentum: consider the whole system (Earth), or the external force(friction).
Proof on law of conservation of momentum:
During collision there’s a force acting on A by B and a force with equal magnitude but opposite direction acting on B by A by Newton’s third law of motion.
FA=-FB, (mAuA - mAvA)/t = (mBvB - mBuB)/t gives the required equation.
Oblique collision in a plane: Assume B is resting and A crashes into B with velocity uA (We assume that A is moving horizontally since we can “turn” the plane.).
After collision, A moves with a velocity vA which makes angle θ with the horizontal, while B move with velocity vB which makes an angle φ with the horizontal. Now let θ be positive (above the horizontal).
Now we decompose the velocity and consider the two direction:
Along x-axis: mAuA = mAvAcosθ + mBvBcosφ
Along y-axis: 0= mAvAsinθ - mBvBsinφ, i.e., mAvAsinθ = mBvBsinφ
Another method is that by mAuA =mAvA+mBvB and they are all vectors, draw a triangle that the angle between mAuA and mAvA is θ and the angle between mAuA and mBvB is φ. By sine law, their relationship is given by mAuA/sin(θ+φ) = mAuA/sin φ = mBuB/sin θ.
Special case: when the two masses are the same and the collision is elastic, the angle θ+φ must be equal to 90 degrees.
Proof: By mAuA+mBuB=mAvA+mBvB , uB=0 and mA=mB, we get uA=vA+vB. (1)
Again since it is an elastic collision we have mAuA2/2+mBuB2/2=mAvA2/2+mBvB2/2 and so that uA2=vA2+vB2. (2)
By (1) we can draw a triangle that adding vA and vB by tip-to-tail method and uA as the sum. Then by (2) and Pythagoras theorem we will get the angle between vA and vB must be 90 degrees.
We call a motion as projectile when the object is given a initial velocity and move under gravitational acceleration (or air resistance as well).
‘When a initial velocity u is given, making angle θ with the horizontal, we will have:
Horizontal: ux=ucosθ, uy=usinθ.
Note that when the vertical velocity is changing due to g, the horizontal velocity WILL NOT CHANGE. (This refers to the Monkey and Hunter experiment.)
Then at an instant t, vx=uxt and vy=uy-gt.
The displacement is given by x=ux and y=uyt-gt2/2.
Now consider some important physical quantities about projectile motion:
1) Max. height: It reaches the highest point when vy=0, i.e., t=uy/g. Substitute t=uy/g into y=uyt-gt2/2 we will have H=ymax=uy2/2g=u2sin2θ/2g.
2) Time of flight: Reaching the highest point and back to the ground takes the same time when we consider a complete projectile motion. Therefore T=2uy/g=2usinθ/g,
3) Range of flight: Obviously R=uxT=2usinθ(ucosθ)/g=u2(2sinθcosθ)/g=u2sin(2θ)/g. That explains that why flying in θ and (90˚-θ) gives the same range.
4) Angle with the horizontal is given by θ’=tan-1(vy/vx)=tan-1(uy-gt/ux).