1. To determine the speed of the 10kg block prior to the collision, we can use the law of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's assume the initial velocity of the 10kg block is v1 and the initial velocity of the 3kg block is v2. The final velocity of both blocks after the collision is 0 since the 3kg block comes to a stop.
The momentum before the collision is given by:
Initial momentum of the 10kg block = mass of the 10kg block * velocity of the 10kg block
= 10kg * v1
Initial momentum of the 3kg block = mass of the 3kg block * velocity of the 3kg block
= 3kg * v2
According to the conservation of momentum:
Initial momentum of the 10kg block + Initial momentum of the 3kg block = Final momentum after the collision
10kg * v1 + 3kg * v2 = 0
Since the 3kg block comes to a stop after the collision, v2 = 0:
10kg * v1 + 3kg * 0 = 0
10kg * v1 = 0
v1 = 0
Therefore, the 10kg block must be at rest prior to the collision in order for the 3kg block to come to a stop after the collision.
2. To determine the spring constant k, we can use the principles of potential energy.
The potential energy of the 3kg block when it is dropped from a height of 1m is given by:
Potential energy = mass * gravitational acceleration * height
= 3kg * 9.8 m/s^2 * 1m
= 29.4 J
The potential energy is converted into the potential energy stored in the compressed spring when the block comes to a stop. The energy stored in a spring is given by:
Potential energy stored in the spring = (1/2) * k * x^2
where k is the spring constant and x is the compression distance.
Plugging in the given values:
29.4 J = (1/2) * k * (10m)^2
29.4 J = 50k
k = 29.4 J / 50
k = 0.588 N/m
Therefore, the spring constant k is 0.588 N/m.