Homework Statement
What is the minimum speed of the car must have at the top of the loop? There are three methods. I would like to know which method or answer is correct.
Please see attached file.
Homework Equations
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The Attempt at a Solution
Please...
No. We could not use Law of conservation of momentum for falling objects because of the external force acting on it. I don't want the answer. I want correct method.
If the time duration of the collision was nonzero, the collision force would have to be specified as a function of time. If it was a nice constant force you could work out the motion during the collision beginning with an F = ma formula for each mass. Looks complicated. The answer would not...
Homework Statement
A ball P of mass m kg is dropped from a point A, which is 2 m vertically above a point B on a horizontal floor. After P hits the floor at B, it rebounds and hits another ball Q, of the same mass, which has also been dropped from A. The impact between the two balls is...
When I calculate the problems for collision,
e=(speed of separation)/(speed of approach)
If two bodies are moving opposite directions, relative velocity is v1+v2.
If two bodies are moving same direction, relative velocity is v2-v1.
After collision two bodies need to separate. It...
You took the wrong value.
We don't know xb and yb
We know centre of mass where is it. x(bar) (or) xc=2, y(bar) or yc=5.5
Use xc=(m1x1+m2x2)/(m1+m2) . You can find x2.
yc formular is similar as xc. You can find y2.
Hello Mandy,
You said subtracted 33 from each side but v1f and v2f are not the same you cannot do like this.
You should use law of conservation of momentum and law of conservation of kinetic energy to solve 2 unknown problem.
You know weight of the rifle. Find mass of the rifle.
Before firing, rifle and bullet haven't moved yet. You can consider their velocities before firing.
I mean after collision, two objects stuck together. So velocities are the same but problem didn't mention direction is angle theta. I think angles are not the same before and after collision.
At the top,direction of normal force and weight are towards the centre and at the bottom, opposite direction. velocity depends on the point for circular track. weight is same as any point but normal force is not the same. At the top normal force value is very small and at the bottom, it's value...
a. F=ma
F=(5.8kg)(9.8m/s2)
F=56.84N
F=-kx
56.84N = -k (.425m - .330m)
k=598.3 N/m or 0.598 kN/m
b. F=-kx
170N = -598.3 N/m x
x= 284.3m
I think part (a) is correct. Part (b) is in correct. Two applied forces are same magnitude and opposite directions. It will cancel each...