Question

(4%) Problem 2: A massless spring (with force constant k = 193 N/m) connects a wall and a block of wood. The system is initially at rest, with the spring unstretched. The block has mass M = 52.3 g and is able to move without friction on a table. A gun is positioned to fire a bullet of mass m = 8.5 g into the block along the spring axis. After the gun is fired, the bullet gets embedded in the block, and the spring is compressed a maximum distance d = 0.83 m. show answer No Attempt 33% Part (a) Find an equation for the speed of the bullet v, just before it hits the block, in terms of the variables given in the problem statement.

Answer 1

Answer:

v = [(m+M)wx^2/m]^1/2

Explanation:

To find the equation for the velocity of the bullet you take into account that just before the bullet hits the block all energy is kinetic energy.

Furthermore, you can assume that all the kinetic energy of the bullet is given to the string. By using this information you have:

$E_k=E_s\\\\\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$v=\sqrt{\frac{kx^2}{m}}\\\\k=(M+m)\omega^2\\\\v=\sqrt{\frac{(M+m)\omega^2 x^2}{m}}$   (this is the equation for the velocity of the bullet)

M: mass of the block = 52.3g

m: mass of the bullet = 8.5g

k: spring constant = 193N/m

x: distance in which the spring is compressed = 0.83m

By replacing in the formula for v you obtain:

v = 125.06m/a

Answer 2

The equation for the speed of the bullet v, just before it hits the block is;

v = kd²/m

We are given;

Mass of block; M = 52.3g

Mass of the bullet; m = 8.5g

Spring constant; k = 193N/m

distance of spring compression; d = 0.83m

By law of conservation of energy, we know that;

Kinetic energy of bullet = kinetic energy of spring. Thus;

¹/₂mv² = ¹/₂kd²

¹/₂ will cancel out to give;

k = mv²/d²  

Making v the subject gives us;

v = kd²/m

Read more at; https://brainly.com/question/13938871