Question

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3. A car with mass 1200 kg travels at 35 m/s, while a second car with mass 1500 kg travels at 25 m/s. The two cars collide in a perfectly inelastic collision.

a) What is the final velocity (speed and direction) if the first car was travelling north and the second travelling south?   
b) If the first car were travelling north and the second were travelling east?

4. Sam stands on a 4-m-long hanging scaffold suspended by two ropes, one at either end.

a) If Sam has a mass of 80 kg, and the scaffold has a mass of 40 kg, what is the minimum tension that could be felt by a single rope if the system is in static equilibrium?
b) What is the maximum tension that could be felt by a single rope if the system is in static equilibrium?

Answer 1

Part A.

When 1st car is traveling in north and 2nd in south (North is +ve y and South is -ve y)

Using momentum conservation:

In perfectly inelastic collision both car will attache and travel together, So

Pi = Pf

m1u1 + m2u2 = (m1 + m2)*V

m1 = mass of 1st car = 1200 kg

m2 = mass of 2nd car = 1500 kg

u1 = +35 m/sec, and u2 = -25 m/sec

V = final velocity of both coupled cars = ?

So,

V = (m1u1 + m2u2)/(m1 + m2)

V = (1200*35 + 1500*(-25))/(1200 + 1500)

V = 1.67 m/sec towards north direction

Part B.

Now since both cars are initially traveling perpendicular with each other, So

Using momentum conservation in x-direction

Pix = Pfx

m1u1x + m2u2x = (m1 + m2)*Vx

u1x = 0 m/sec = initial velocity of m1 in x-direction (in east)

u2x = 25 m/sec = initial velocity of m2 in x-direction

Vx = final velocity of Coupled cars in x-direction = ?

Vx = (m1u1x + m2u2x)/(m1 + m2)

Vx = (1200*0 + 1500*25)/(1200 + 1500) = 13.89 m/sec

Now, Using momentum conservation in y-direction

m1u1y + m2u2y = (m1 + m2)*Vy

u1y = 35 m/sec = initial velocity of m1 in y-direction (towards north)

u2y = 0 m/sec = initial velocity of m2 in y-direction

Vy = final velocity of Coupled skaters in y-direction = ?

Vy = (m1u1y + m2u2y)/(m1 + m2)

Vy = (1200*35 + 1500*0)/(1200 + 1500) = 15.56 m/sec

to find magnitude of velocity: Square and add both equation and take square root

V = sqrt (Vx^2 + Vy^2)

V = sqrt (13.89^2 + 15.56^2)

V = 20.86 m/sec

to find direction of final velocity:

Direction = arctan (Vy/Vx)

= arctan (15.56/13.89) = 48.25 deg

So final velocity = 20.86 m/sec at 48.25 deg North of East

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