Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m

The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:

The velocity of the ball is 4.64m/s
(c) The acceleration is given by:


The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:

THe compression of the ball when it strikes the floor is 5.11*10^-3m
I think it false. Sorry if i'm wrong.
Answer:
The magnitude of the force is 12 N Upwards
Explanation:
The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.
Given;
magnitude of charge, q = 0.06 C
magnitude of electric field, E = 200 N/C
The magnitude of the force is given by;
F = qE
F = 0.06 x 200 N/C
F = 12 N Upwards
Therefore, the magnitude of the force is 12 N Upwards