Answer:
Canyon is 50.176 meter deep.
Explanation:
The students is standing on the rim of the canyon and drops down a rock from the rim(cliff). We have to find what is the depth of the canyon i.e. how much below is the ground from the cliff.
Given data:
Time = t = 3.2 s
Initial velocity = 
= 0 m/s
Gravitational acceleration = g = 9.8 m/s²
Height = h = ?
According to second equation of motion
As initial velocity is zero, So the first term of right hand side of above equation equal to zero
h = (0.5)(9.8)(3.2)²
h = 50.176 m
This means, the rock traveled 50.176 meters to reach the bottom of the Canyon. So, the Canyon is 50.176 meter deep.
Momentum is conserved.
p = m₁v₁ + m₂v₂ = constant
Momentum before the collision(kick):
p = 80 * 0 + 4 * 0 = 0
Momentum after the collision:
p = 80 * v₁ + 4 * 15 = 0
Solve for v₁.
k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
