Inelastic Collision from different but not opposite directions. HELP

AI Thread Summary
In a completely inelastic collision, two skaters collide and move together post-collision. Alfred, weighing 83 kg and moving east at 6.3 km/hr, collides with another skater weighing 55 kg, moving north at 7.8 km/hr. To find the final velocity after the collision, momentum must be treated as a vector, requiring the use of the equation M1V1 + M2V2 = (M1 + M2)V'. The velocities should be broken down into their respective components due to their different directions. This approach will guide the solution to the problem effectively.
A PP4L K
Messages
2
Reaction score
0

Homework Statement


Two skaters collide and embrace, in a completely inelastic collision. Alfred, whose mass is 83 kg is originally moving east with a speed of 6.3km/hr. The other skater (m=55 kg) is originally moving north with a speed of 7.8km/hr. What is the velocity (magnitude and direction) after the collision?


Homework Equations


1)M1V1 + M2V2 = (M1 + M2)V’
2)F∙Δt = M∙Vf - M∙Vi
3)P = MV
4)Ft = Δp
5)I = MVF - MVi

The Attempt at a Solution


M1= 83kg
V1= 1.75 m/s
M2= 55kg
V2= 2.17 m/s


Honestly I have no idea where to even start. I know how to do regular inelastic, and elastic, collision problems. This is an extra credit problem that I am trying to do, so if anyone could show me how to work one of these out or could hint me in the right direction it would be very much appreciated! :)
 
Physics news on Phys.org
Okay. So you start off with the equation M1V1 + M2V2 = (M1+M2)V'. Treat momentum as a vector and solve for it in components.
 
keep in mind that the velocities of each skater are on different axes
like ben.tien said, use that equation M1V1 + M2V2 = (M1 +M2)V and add the velocities as vectors in components. That should get you on the right path
 
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}...
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...
Back
Top