Answer:
V = 5.83 m/s
Explanation:
Given that,
Mass of a ball of a clay, m = 2 kg
Initial speed of the clay, u = 35 m/s
Mass of a box, m' = 10 kg
Initially, the box was at rest, u' = 0
We need to find the velocity of the box after the collision. Let V be the common speed. Using the conservation of momentum to find it.
So, the velocity of the box after the collision is equal to 5.83 m/s.
(a) 3.5 Hz
The angular frequency in a spring-mass system is given by
where
k is the spring constant
m is the mass
Here in this problem we have
k = 160 N/m
m = 0.340 kg
So the angular frequency is
And the frequency of the motion instead is given by:
(b) 0.021 m
The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at
x = A
where A is the amplitude of the motion.
The maximum displacement is given by Hook's law:
where
F is the force applied initially to the spring, so it is equal to the weight of the block:
k = 160 N/m is the spring constant
Solving for A, we find