A bullet with mass 4.91 g is fired horizontally into a 1.981-kg block attached to a horizontal spring. The spring has a constant 6.25 102 N/m and reaches a maximum compression of 6.38 cm.


a) find the initial speed of the bullet-block system.
b) find the speed of the bullet

a) By "initial speed", they really mean "right after the bullet hits the block and gets embedded". Call that speed v. The bullet-block speed is zero before that.

Assume energy is conserved during spring compression.

(1/2)(M+m) v^2 = (1/2)k*X^2
v = sqrt [k/(M+m)]*X
= sqrt[625/1.985] s^-1 *0.0638 m
= 1.132 m/s

b) For the initial bullet speed V, apply conservation of momentum to the bullet-stopping process. Mechanical energy is NOT conserved then.
m*V = (m+M)*v

(0.00491 kg)* V = (1.985 kg)* v

V = 458 m/s

a) Well, this sounds like a real spring-loaded situation! To find the initial speed of the bullet-block system, we can start by calculating the potential energy stored in the compressed spring when it reaches maximum compression. This potential energy can be equated to the kinetic energy that the bullet imparts to the block.

Using the formula for potential energy in a spring, we have:

Potential energy = (1/2) * k * x^2

Where k is the spring constant and x is the maximum compression.

Plugging in the values we have, the potential energy is:

Potential energy = (1/2) * (6.25 * 10^2 N/m) * (0.0638 m)^2

Now, since the potential energy is also equal to the kinetic energy, we can solve for the initial speed of the bullet-block system:

(1/2) * m_bullet * v_initial^2 = (1/2) * k * x^2

Simplifying the equation, we have:

m_bullet * v_initial^2 = k * x^2

Plugging in the values we have, the equation becomes:

(4.91 * 10^-3 kg) * v_initial^2 = (6.25 * 10^2 N/m) * (0.0638 m)^2

Now, all we need to do is solve for v_initial:

v_initial^2 = [(6.25 * 10^2 N/m) * (0.0638 m)^2] / (4.91 * 10^-3 kg)

v_initial^2 ≈ 4.078 m^2/s^2

Taking the square root of both sides, we find that the initial speed of the bullet-block system is approximately 2.019 m/s.

b) Now, to find the speed of the bullet itself, we need to consider the conservation of momentum before and after the collision. Since the block and bullet are initially at rest, the total momentum of the system before the collision is zero.

After the bullet is fired into the block, the two masses will move together with a common velocity. Let's call this final velocity V_combined.

Since momentum is conserved, we have:

(m_bullet * v_initial) + (m_block * 0) = (m_bullet + m_block) * V_combined

Plugging in the values we have, the equation becomes:

(4.91 * 10^-3 kg) * 2.019 m/s = (4.91 * 10^-3 kg + 1.981 kg) * V_combined

Now, all we need to do is solve for V_combined:

(4.91 * 10^-3 kg) * 2.019 m/s = (4.91 * 10^-3 kg + 1.981 kg) * V_combined

V_combined ≈ 0.010 m/s

So, the speed of the bullet after being fired into the block is approximately 0.010 m/s.

Clown Bot hopes this spring-loaded explanation didn't spring any surprises, and that it brought a smile to your face!

a) To find the initial speed of the bullet-block system, we can use the conservation of momentum. The momentum before the collision is equal to the momentum after the collision.

The momentum before the collision can be calculated as the mass of the bullet times its initial velocity (since the block is initially at rest):
m_bullet * v_initial = (0.00491 kg)(v_initial)

The momentum after the collision is the sum of the momenta of the bullet and the block:
m_bullet * v_bullet + m_block * v_block

Since the bullet is fired horizontally, its vertical velocity is 0. Therefore, we can substitute v_bullet = v_initial into the equation.

Now, we can rearrange the equation to solve for v_initial:
m_bullet * v_initial = m_bullet * v_bullet + m_block * v_block
v_initial = (m_bullet * v_bullet + m_block * v_block) / m_bullet

Substituting the known values:
v_initial = (0.00491 kg * v_bullet + 1.981 kg * 0) / 0.00491 kg
v_initial = v_bullet

Therefore, the initial speed of the bullet-block system is equal to the speed of the bullet.

b) To find the speed of the bullet, we can use the conservation of mechanical energy. The mechanical energy is conserved throughout the motion, which includes the bullet's motion before and after the collision.

The initial mechanical energy of the system is equal to the final mechanical energy of the system.

We can calculate the initial mechanical energy:
E_initial = 1/2 * m_bullet * v_initial^2 + 1/2 * k * x^2,
where k is the spring constant and x is the maximum compression of the spring.

The final mechanical energy is the kinetic energy of the bullet after the collision:
E_final = 1/2 * m_bullet * v_bullet^2.

Since the mechanical energy is conserved:
E_initial = E_final,
1/2 * m_bullet * v_initial^2 + 1/2 * k * x^2 = 1/2 * m_bullet * v_bullet^2.

Rearranging the equation to solve for v_bullet:
v_bullet^2 = (v_initial^2 + (k / m_bullet) * x^2).

Substituting the known values:
v_bullet^2 = (v_initial^2 + (625 N/m / 0.00491 kg) * (0.0638 m)^2).
v_bullet^2 = (v_initial^2 + 81291 m^2/s^2),
v_bullet = sqrt(v_initial^2 + 81291 m^2/s^2).

Note: Since the initial speed of the bullet-block system is equal to the speed of the bullet, we can substitute v_initial for v_bullet in the final equation.

Therefore, the speed of the bullet is given by:
v_bullet = sqrt(v_initial^2 + 81291 m^2/s^2).

To find the initial speed of the bullet-block system, we can use the principle of conservation of linear momentum. The linear momentum before the collision is equal to the linear momentum after the collision.

The linear momentum is given by the product of mass and velocity: momentum = mass × velocity.

Let's assume the velocity of the bullet before the collision is v. Since the bullet is fired horizontally into the block, the velocity of the block before the collision is 0.

Before the collision:
Momentum of bullet = (mass of bullet) × (velocity of bullet)
= (4.91 g) × v

After the collision, the bullet and block move together with a common velocity, which we can call V. So the momentum after the collision is:
Momentum of bullet + block = (mass of bullet + mass of block) × (velocity of bullet + velocity of block)
= (4.91 g + 1.981 kg) × V

According to the principle of conservation of linear momentum:
Momentum before collision = Momentum after collision

Therefore, we have:
(4.91 g) × v = (4.91 g + 1.981 kg) × V

Now we can solve for v, the initial speed of the bullet-block system.

a) To find the initial speed of the bullet-block system, use the equation:
v = [(4.91 g + 1.981 kg) × V] / (4.91 g)

Now let's move on to part b, finding the speed of the bullet.

b) After the collision, the bullet gets embedded in the block, and they move together. To find the speed of the bullet, we need to find the final velocity of the bullet-block system after the collision.

We can use the conservation of mechanical energy to find the final velocity. The energy stored in the spring when it is compressed is equal to the kinetic energy of the bullet and block system.

The potential energy stored in the spring is given by:
Potential energy = (1/2) × (spring constant) × (maximum compression)^2

The kinetic energy of the bullet and block system is given by:
Kinetic energy = (1/2) × (mass of bullet + mass of block) × (final velocity)^2

According to the conservation of mechanical energy:
Potential energy = Kinetic energy

Thus, we have:
(1/2) × (6.25 × 10^2 N/m) × (0.0638 m)^2 = (1/2) × (4.91 g + 1.981 kg) × (final velocity)^2

Now we can solve for the final velocity, which will give us the speed of the bullet.

b) To find the speed of the bullet, use the equation:
final velocity = sqrt([(6.25 × 10^2 N/m) × (0.0638 m)^2] / (4.91 g + 1.981 kg))

By calculating the values in the equations, you will get the answers to both parts a) and b).