Impulse and Momentum: A Comprehensive Guide with Solved Problems, Lecture notes of Engineering Mathematics

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IMPULSE & MOMENTUM

Impulse (J) :

Refers to the product of the force and the time

interval it acts on the body.

J = F(t

- t

) = F(Δt)

where : F – applied constant force ( in Newtons) Δt – time interval (in seconds) J – Impulse ( in Ns) F m m v 1 v 2 t 1 t 2

Both Momentum & Impulse are vector quantities, thus they have both horizontal & vertical components. And follow standard sign conventions X - component

p

x

= mv

x J

x

= F

x

(t

- t

) = F

x

(Δt)

Y - component

p

y

= mv

y J

y

= F

y

(t

- t

) = F

y

(Δt)

p = p

x

+ p

y

J = J

x

+ J

y

F m m v 1 v 2 t 1 t 2 Relationship between Momentum & Impulse J = Δp FΔt = mv 2 – mv 1 “The change in momentum at any time interval equals the impulse of the force applied during that time interval.” FxΔt = mv2x – mv1x FyΔt = mv2y – mv1y

2. Prob. 8 - 12 ] A bat strikes a 0. 145 kg baseball. Just before impact, the ball is traveling horizontally to the right at 50 m/s, and it leaves the bat traveling to the left at an angle of 30 ° above the horizontal with a speed of 65 m/s. If the ball and bat are in contact for 1. 75 msec, find the horizontal & vertical components of the average force on the ball.

SAMPLE PROBLEMS

v 1 = 50 m/s Fx Fy F 30 ° x-component : vectors to the right (+) FxΔt = m(v2x – v1x) Fx = m(v2x – v1x) / Δt Fx = (0.145kg) [(-65m/s)(cos30°) – (+50m/s)] / (0.00175 s) F x = - 8,807 N y-component : vectors going up (+) FyΔt = m(v2y – v1y) Fy = m(v2y – v1y) / Δt Fy = (0.145kg) [(+65m/s)(sin30°) – 0] / (0.00175 s) F y = +2,692.9 N

3. Prob. 8 - 1 ] (a) What is the magnitude of the momentum of a 10 , 000 kg truck whose speed is 12 m/s? (b) What speed would a 2 , 000 kg SUV have to attain in order to have i. The same momentum (as the truck) ii. The same kinetic energy (as the truck)

SAMPLE PROBLEMS

1. An open-topped freight car with a mass of 10 , 000 kg is coasting without friction along a level track. It is raining very hard, and the rain is falling vertically downward. The car is originally empty and moving with a speed of 3 m/s. What is the speed of the car after it has traveled long enough to collect 1 , 000 kg of rainwater? m 1 v 1 + m 2 v 2 = m 1 u 1 +m 2 u 2 Let: m 1 = mass of freight car = 10,000 kg

SAMPLE PROBLEMS

Given : Required :ucar m 1 = 10,000 kg v 1 = 3 m/s m 2 = 1,000 kg u 1 =? m 2 = mass of rain water = 1,000 kg m 1 v 1 + m 2 v 2 = m 1 u 1 +m 2 u 2 (10,000 kg)(3m/s )+(0)(0 m/s) = (10,000 kg)u 1 +(1,000 kg)u 2 30,000 kg-m/s = (10,000 kg)u 1 +(1,000 kg)u 2 By logic u 1 = u 2 , since rain water moving along with the car 30,000 kg-m/s = (10,000 kg)u 1 +(1,000 kg)u 1 30,000 kg-m/s = u 1 (10,000kg+1,000 kg) 30,000 kg-m/s = u 1 (11,000kg) (30,000 kg-m/s)/11,000kg = u 1 2.727 m/s = u 1 1 2

mAvA + mBvB = mAuA+mBuB

SAMPLE PROBLEMS

2. Given : Required :uA (1kg)(0 m/s)+(2kg)(0 m/s) = (1kg)uA+(2kg)(0.9 m/s) uA = 1.8 m/s to the left 1 2 A mA = 1kg mB = 2 kg B A uA =? uB = 0.90 m/s B 0 = (1kg)uA+1.8kg-m/s **- 1.8 kg-m/s = (1kg)uA (- 1.8 kg-m/s)/1kg = uA

• 1.8 m/s = uA**

ELASTIC & INELASTIC COLLISIONS COLLISIONS

- Defined as a kind of interaction between two bodies wherein there is a strong interaction that last for a relatively short time. - Usually treated as an isolated system because the interactive forces are greater than the external forces (i.e. friction).

ELASTIC & INELASTIC COLLISIONS TYPES OF COLLISIONS

1. Perfectly Elastic Collision A B A B vA uB uA vB Before and During Collision After Collision

ELASTIC & INELASTIC COLLISIONS TYPES OF COLLISIONS

2. Perfectly Inelastic Collision A B A B vA u vB Before and During Collision After Collision

ELASTIC & INELASTIC COLLISIONS TYPES OF COLLISIONS

2. Perfectly Inelastic Collision After collision, the colliding bodies merge and move in one direction at with a common velocity. Inelastic collisions don't conserve kinetic energy, but total momentum before and after collision is conserved. Kinetic energy after collision is less than that before collision. m A v A + m B v B = u (m A + m B )

Examples

_2. A toy car with mass of 0. 3 kg moves to the right at 5 m/s along a frictionless horizontal table and collides with a toy truck having a mass of 0. 8 kg which is moving

1. 5 m/s to the left. If the two toys stick together, what is the final velocity (magnitude & direction)? Given : Required : u_ mcar = 0.3 kg vcar = 5 m/s mtrk = 0.8 kg vtrk = 1.5 m/s u =? Before : After :

Examples

3. A 2000 kg automobile going eastward on Ortigas Ave at 50 km/hr collides with a 4000 kg truck which is going northward across Ortigas Ave at 20 km/hr. If they become coupled on collision, what is the magnitude and direction of their velocity immediately after collision? (Friction forces between the cars & the road can be neglected during the collision). Given : Required : u vcar = 50 km/hr vtrk = 20 km/hr u =? Before : (^) After :