Momentum is conserved after the collision
Momentum of 2 Kg before collision = 2 * 3 = 6
Momentum of 4 kg before collision = 4 * -3 = -12
so 6 + -12 = 2 * -4 + 4 *x where x is velocity of 4kg marble.
4x - 8 = -6
4x = 2
x = 0.5
Velocity of 4 kg marble is 0.5 m/s after collision
The 2 kg marble will move in the opposite direction to which it was moving before the collision.
Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
Substituting all the values in above equation,
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answer:
The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.
It is a comet that was a comet
To what i see, the answer is....
C.
