Mechanics problem: momentum conservation..EMERGENCY!!?

Question

Answer with explanation please :/

Alan · Accepted Answer

Part a)

The total momentum (mass times velocity) is always conserved, but in this case their are 2 momentums, one in the x direction and one in the y.

Mx = -(1170/9.8)*69 = 8238 kg*m/s left

My = -(1760/9.8)*49 = 8800 kg*m/s down

To find the magnitude you use Pythagorean's Theorem because the x and y components form a right triangle.

Mx^2 + My^2 = Mf^2Mf = 12054 kg*m/s

Now that you know the final momentum you can divide by the total mass to find the velocity.

(1170+1760)/9.8 = 299 kg

Mf/299 = 40.3 m/s

Part b)

To find the direction you will have to use Tan^-1 on the x and y components, so Tan^-1(My/Mx) = angle = 46.9 degrees

Because the cars are going left and down, the resultant will be in the third quadrent so the angle you got is the angle counter-clockwise from the negative x axis, meaning you will have to add 180 degrees to your answer.

46.9+180 = 226.9 degrees from East

oubaas · Answer

Momentum conservation shall apply :
V = (1170*69+1760*49)/(1170+1760) = 57.0 m/sec

Ash · Answer

m₁ = (1170 / 9.81) kg

u₁ = 69 m/s in West direction
m₂ = (1760 / 9.81) kg
u₂ = 49 m/s in South direction
v = ?

momentum conservation in East-West direction
m₁u₁ + m₂u₂ = (m₁+m₂)v
v(east-west) = (m₁u₁ + m₂u₂) / (m₁+m₂)
v(east-west) = ( (1170 / 9.81)69 + (1760 / 9.81) 0) / ((1170 / 9.81)+ (1760 / 9.81))
v(east-west) =  [(1170*69)/9.81 ] / [(1170 + 1760)/9.81]
v(east-west) = 80730 / 2930
v(east-west) =  27.6 m/s

momentum conservation in North-South direction
v(north-south) = (m₁u₁ + m₂u₂) / (m₁+m₂) 
v(north-south) = ( (1170 / 9.81)0 + (1760 / 9.81) 49) / ((1170 / 9.81)+ (1760 / 9.81))
v(north-south) = 86240 / 2930
v(north-south) = 29.4 m/s

Resultant velocity, v = √(27.6² + 29.4²) = 40.3 m/s

angle S of W = tan⁻¹ (29.4/27.6) = 46.8°

With respect to East, the angle will be 180°+46.8 = 226.8°

Mechanics problem: momentum conservation..EMERGENCY!!?

3 Answers