LINEAR MOMENTUM

Direction of Linear momentum will be just same as direction of velocity

Case I Changing of linear momentum

Note: (1) Slope of p - t graph will give us Force (F)

(2) Area under the F - t carve will gives us change in momentum

Impulse FDt =Dp Also we can say impulse is change in momentum

Example: A ball of mass 1 kg drops on the hard floor and ball strikes the floor with 10m/s rebound with same speed also. If ball remains for 0.1 sec in contact with floor. Then find

(i) Magnitude of rate of change of momentum

(ii) Rate of change of magnitude of momentum

Solution: Momentum just before strike the floor p1 = mv

p1 = - 1 x10

= - 10 kg mt/sec

Momentum just after strike the floor

p2 = + mv

= + 1 x10

= 10 kg mt/sec

(i) Magnitude of rate change of momentum

(ii) Rate of change of magnitude of momentum

Þ Magnitude of momentum is not changing only direction of momentum is changing

Þ Therefore                                                  Ans-Zero

CHANGING IN MOMENTUM

Change in momentum along x- axis Þ

p1 = mv sinq before Collision

p2 = mv sinq after Collision

(D px) = p2p1 = 0

Change in momentum along y-axis Þ

p1 = - mv cosq before Collision

p2 = mv cosq after Collision

(Dp)y = p2- p1 = 2 mv cosq

CHANGING IN MOMENTUM

Let assume a ball attached with a string and moving uniformly in a Circular path

(i) Change in momentum from A to B

(ii)  Change in momentum from A to C

i

Another case

But here

v2 = u2 – 2gl

Case I Conservation of linear momentum (COLM)

Example:

Given A trolly of mass M kept at rest on a Smooth road. Two person each of mass m Standing at two ends of trolly. Now person A jumps in horizontal  direction form the trolly with speed u (with respect to trolly) and then person B jumps in horizontal direction with speed u (with respect to trolly) in opposite direction. Find final speed of trolly just after both person jumps from the trolly.

Solution

Just before A jumps

p1 = (2 m + m) x 0 = 0

Just after A jumps

p2 = mu + (M + m) v1

Now  Fext = 0, applying COLM

p1 = p2 Þ mu + (M + m) v1 = 0

Again if we assume that trolly and person B is at rest

Just before B jumps

p1 = (M + m) x 0 (Assume)

Just after B jumps

p2 = - mu + Mv2

Now applying COLM

p1 = p2Þ - mu + Mv2 = 0

Now Final speed of trolly after both A & B Jumps

Vf = V1 + V2

COLLISION

OBLIQUE COLLISION

Types of Collisions

(1) Perfect  elastic Collision.

Momentum will Conserved p1 = p2

Energy will Conserved k1 = k2

No loss of energy

Coefficient of restitution e = 1

CONCEPT OF EXCHANGE

(2) Elastic Collision

Momentum will Conserved p1 = p2

Energy will not Conserved k1¹ k2

Minimum energy loss  Coefficient of restitution

e < 1 (But here e will be Close to 1)

(3) Perfect inelastic Collision

Momentum will conserved p1 = p2

Energy will not conserved k1 ≠ k2

Maximum energy loss coefficient of restitution e = 0

(4) Inelastic Collision

Momentum will Conserved p1 = p2

Energy will not Conserved k1 ¹ k2

Maximum energy loss Coefficient of restitution

e > 0 (But here e will be close to 0)

NEWTONS LAW FOR COLLISION

Speed just after Collision = e (Speed just before Collision)

Height after Collision = e2 (Height before Collision)

Total distance Covered by the ball

H = h + 2 e2 h + 2e4 h + -------

= h [ 1 + 2e2 (1 + e2 + e4 + ----)]