Momentum before the hit:
p = mv = 0.01 * 300 + 1 * 0
Momentum after the hit:
p = 0.01 * 150 + 1 * v
Momentum is conserved:
0.01 * 300 = 0.01 * 150 + v
3 = 1.5 + v
v = 1.5
The velocity of the block after the collision is 1.5 m/s.
Answer:
An object moving in certain direction with an acceleration in the perpendicular direction. The above condition is possible . Example of such situation in life would be when stone tied to a string whirling in a circular path
Hope this helps and pls mark as BRAINLIEST :)
The magnitude of the unknown height of the projectile is determined as 16.1 m.
<h3>
Magnitude of the height</h3>
The magnitude of the height of the projectile is calculated as follows;
H = u²sin²θ/2g
H = (36.6² x (sin 29)²)/(2 x 9.8)
H = 16.1 m
Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.
Learn more about height here: brainly.com/question/1739912
#SPJ1
Complete Question:
A 10 kg block is pulled across a horizontal surface by a rope that is oriented at 60° relative to the horizontal surface.
The tension in the rope is constant and equal to 40 N as the block is pulled. What is the instantaneous power (in W) supplied by the tension in the rope if the block when the block is 5 m away from its starting point? The coefficient of kinetic friction between the block and the floor is 0.2 and you may assume that the block starting at rest.
Answer:
Power = 54.07 W
Explanation:
Mass of the block = 10 kg
Angle made with the horizontal, θ = 60°
Distance covered, d = 5 m
Tension in the rope, T = 40 N
Coefficient of kinetic friction, 
Let the Normal reaction = N
The weight of the block acting downwards = mg
The vertical resolution of the 40 N force, 





Power, 
