Answer:
the magnitude of the displacement after 5s is 137.31 m.
Explanation:
Given;
initial velocity of the projectile, u = 60 m/s
angle of projection, θ = 60°
time of motion, t = 5s
the vertical component of the velocity,
The magnitude of the displacement after 5s is calculated as;
Therefore, the magnitude of the displacement after 5s is 137.31 m.
Answer:
Velocity of the two balls after collision: .
of kinetic energy would be lost.
Explanation:
<h3>Velocity</h3>Because the question asked about energy, convert all units to standard units to keep the calculation simple:
The two balls stick to each other after the collision. In other words, this collision is a perfectly inelastic collision. Kinetic energy will not be conserved. The velocity of the two balls after the collision can only be found using the conservation of momentum.
Assume that the system of the two balls is isolated. Thus, the sum of the momentum of the two balls will stay the same before and after the collision.
The momentum of an object of mass and velocity
is:
.
Momentum of the two balls before collision:
Based on the assumptions, the sum of the momentum of the two balls after collision should also be . The mass of the two balls, combined, is
. Let the velocity of the two balls after the collision
. (There's only one velocity because the collision had sticked the two balls to each other.)
These two values are supposed to describe the same quantity: the sum of the momentum of the two balls after the collision. They should be equal to each other. That gives the equation about :
.
.
In other words, the velocity of the two balls right after the collision should be .
The kinetic energy of an object of mass and velocity
is
.
Kinetic energy before the collision:
The two balls stick to each other after the collision. Therefore, consider them as a single object when calculating the sum of their kinetic energies.
Sum of the kinetic energies of the two balls right after the collision:
.
Therefore, of kinetic energy would be lost during this collision.