A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? Use conservation of momentum.
After the bullet embeds inside the block, is it going to be m1+m2?
Now I get it, thanks.
To find the distance the spring is compressed, we can use the principle of conservation of momentum. According to this principle, the momentum before the collision should be equal to the momentum after the collision.
Let's break down the problem into two parts: the bullet's motion before it hits the block and the combined motion of the block and bullet after the collision.
1. Bullet's motion before the collision:
The mass of the bullet, m1 = 10 g = 0.01 kg
The initial velocity of the bullet, u1 = 300 m/s
The momentum before the collision can be calculated using the formula: momentum = mass * velocity
Momentum before the collision, p1 = m1 * u1
2. Combined motion of the block and bullet after the collision:
The combined mass of the block and bullet, m2 = 2 kg + 0.01 kg = 2.01 kg
Let's assume that after the collision, the block and bullet move together with a final velocity, v2.
The momentum after the collision can be calculated using the formula: momentum = mass * velocity
Momentum after the collision, p2 = m2 * v2
According to the conservation of momentum principle, the momentum before the collision should be equal to the momentum after the collision.
p1 = p2
Therefore, m1 * u1 = m2 * v2
Now, let's substitute the known values into the equation and calculate v2:
0.01 kg * 300 m/s = 2.01 kg * v2
v2 = 0.01 kg * 300 m/s / 2.01 kg
v2 ≈ 0.149 m/s
Since the block and bullet move together as a single mass, we can consider this as a case of an elastic collision. In an elastic collision, both momentum and kinetic energy are conserved.
Using the conservation of kinetic energy, we can write the equation:
(1/2) * m1 * u1^2 = (1/2) * m2 * v2^2
Simplifying the equation:
0.01 kg * (300 m/s)^2 = 2.01 kg * v2^2
v2^2 ≈ [0.01 kg * (300 m/s)^2] / 2.01 kg
v2^2 ≈ 448.256 m^2/s^2
v2 ≈ 21.171 m/s (approx.)
Next, we will calculate the compression distance of the spring.
The potential energy of a spring can be given by the equation:
Potential energy = (1/2) * k * x^2
Where k is the force constant of the spring and x is the compression distance.
The work done by the bullet in compressing the spring is equal to the potential energy stored in the spring.
The work done by the bullet can be calculated using the work-energy principle:
Work done = Change in kinetic energy
Work done = (1/2) * m2 * v2^2 - 0
Since there is no initial velocity of the block, the initial kinetic energy is zero.
Therefore,
Work done = (1/2) * m2 * v2^2
Now, let's substitute the known values into the equation:
Work done = (1/2) * 2.01 kg * (21.171 m/s)^2
Work done ≈ 449.108 J
The work done is equal to the potential energy stored in the spring:
Potential energy = Work done
(1/2) * k * x^2 = 449.108 J
Substituting the force constant of the spring, k = 19.6 N/m:
(1/2) * 19.6 N/m * x^2 = 449.108 J
Now, solve for x:
9.8 N/m * x^2 = 449.108 J
x^2 ≈ 449.108 J / 9.8 N/m
x^2 ≈ 45.919 m^2
Therefore,
x ≈ √45.919 m
x ≈ 6.77 m (approx.)
Hence, the spring is compressed approximately 6.77 meters.
Conservation of momentum tells you the velocity V' of the block and embedded bullet before the spring gets compressed.
0.010 kg*300 m/s = V' * (2.01 kg)
Solve for V'.
Energy is conserved during spring compression.
(1/2) k X^2 = (1/2) (M+m) V'^2
k is the spring force constant. M is block mass. m is bullet mass.
Solve for spring compression X