Solving Momentum Collision Problems

  • Thread starter Thread starter Ginerva123
  • Start date Start date
  • Tags Tags
    Collision Momentum
AI Thread Summary
The discussion revolves around solving a momentum collision problem involving two objects with different masses and velocities. The initial calculations for the velocities prior to the collision were based on the momentum conservation principle, but the user received incorrect answers. A key mistake identified was the misapplication of sine and cosine functions due to misunderstanding the angle's orientation, as 24 degrees east of north indicates a greater northward direction than eastward. Correcting this misunderstanding is crucial for accurately determining the initial velocities. The conversation emphasizes the importance of vector components in momentum calculations.
Ginerva123
Messages
13
Reaction score
0
[SOLVED] Momentum - collision

Homework Statement



An object of mass 950kg traveling east collides with an object of mass 1900kg traveling north. The two objects then move off together at 16m/s in a direction of 24degrees east of north. Find their respective velocities prior to the collision.

Homework Equations



p = mv

The Attempt at a Solution


Momentum after collision = 2850 x 16 = 45600
As momentum is a vector, then breaking it into its components we get:
Object 1: 950 x v1 = (45600)cos24 ... v1=43.85m/s
Object 2: 1900 x v2 = (45600)sin 24... v2=9.76m/s

I thought this seemed simple enough, but apparently the answers are wrong... can anyone see where I might have made mistake, or am I just going about it the wrong way altogether?
 
Physics news on Phys.org
Ginerva123 said:

Homework Statement



An object of mass 950kg traveling east collides with an object of mass 1900kg traveling north. The two objects then move off together at 16m/s in a direction of 24degrees east of north. Find their respective velocities prior to the collision.

Homework Equations



p = mv

The Attempt at a Solution


Momentum after collision = 2850 x 16 = 45600
As momentum is a vector, then breaking it into its components we get:
Object 1: 950 x v1 = (45600)cos24 ... v1=43.85m/s
Object 2: 1900 x v2 = (45600)sin 24... v2=9.76m/s

I thought this seemed simple enough, but apparently the answers are wrong... can anyone see where I might have made mistake, or am I just going about it the wrong way altogether?

you got the sin/cos mixed up... 24 degrees "east of north" is pointing more north than east...
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top