Answer: 909 m/s
Explanation:
Given
Mass of the bullet, m1 = 0.05 kg
Mass of the wooden block, m2 = 5 kg
Final velocities of the block and bullet, v = 9 m/s
Initial velocity of the bullet v1 = ? m/s
From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve
m1v1 = (m1 + m2) v, on substituting, we have
0.05 * v1 = (0.05 + 5) * 9
0.05 * v1 = 5.05 * 9
0.05 * v1 = 45.45
v1 = 45.45 / 0.05
v1 = 909 m/s
Thus, the original velocity of the bullet was 909 m/s
Answer:
The potential energy (P.E) at the top is 392 J
The kinetic energy (K.E) at the top is 0 J
The potential energy (P.E) at the halfway point is 196 J.
The kinetic energy (K.E) at the halfway point is 196 J.
Explanation:
Given;
mass of the rock, m = 2 kg
height of the cliff, h = 20 m
speed of the rock at the halfway point, v = 14 m/s
The potential energy (P.E) and kinetic energy (K.E) when its at the top;
P.E = mgh
P.E = (2)(9.8)(20)
P.E= 392 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the top of the cliff = 0
K.E = ¹/₂(2)(0)²
K.E = 0
The potential energy (P.E) and kinetic energy (K.E) at the halfway point;
P.E = mg(¹/₂h)
P.E = (2)(9.8)(¹/₂ x 20)
P.E = 196 J
K.E = ¹/₂mv²
where;
v is velocity of the rock at the halfway point = 14 m/s
K.E = ¹/₂(2)(14)²
K.E = 196 J.
Answer:
a
an apple falling from a tree