A 0.02kg bullet collides with a 5.75kg pendulum. After the collision, the pair swings up to a maximum height of 0.386m. Determine the velocity of the bullet just before impact.
Let the velocity of the (bullet+pendulum) is v m/s after the collision,
=>By the law of energy conservation:-
=>PE(final) = KE(initial)
=>mgh = 1/2mv^2
=>v = √2gh
=>v = √[2 x 9.8 x 0.386]
=>v = 2.75 m/s
Let the velocity of the bullet before the impact was u m/s, By the law of momentum conservation:-
=>m1u1+m2u2 = (m1+m2)v
=>0.02 x u + 0 = (0.02+5.75) x 2.75
=>u = 793.38 m/s
To determine the velocity of the bullet just before impact, we can use the principles of conservation of momentum.
The equation for conservation of momentum is:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Where:
m1 = mass of the bullet (0.02 kg)
m2 = mass of the pendulum (5.75 kg)
v1 = initial velocity of the bullet (unknown)
v2 = initial velocity of the pendulum (initially at rest, so 0 m/s)
v1' = final velocity of the bullet (unknown)
v2' = final velocity of the pendulum (unknown)
Since the pendulum is initially at rest, we can simplify the equation to:
m1 * v1 = m1 * v1' + m2 * v2'
Substituting the values we know:
0.02 kg * v1 = 0.02 kg * v1' + 5.75 kg * 0.0 m/s
Simplifying further:
0.02 kg * v1 = 0.02 kg * v1'
Now, we need to consider the conservation of mechanical energy. The equation for conservation of mechanical energy is:
1/2 * m1 * v1^2 + 1/2 * m2 * v2^2 = m1 * g * h
Where:
v1 = initial velocity of the bullet (unknown)
v2 = initial velocity of the pendulum (0 m/s)
g = acceleration due to gravity (9.8 m/s^2)
h = maximum height reached by the pendulum (0.386 m)
Substituting the values we know:
1/2 * 0.02 kg * v1^2 + 1/2 * 5.75 kg * 0^2 = 0.02 kg * 9.8 m/s^2 * 0.386 m
Simplifying further:
1/2 * 0.02 kg * v1^2 = 0.02 kg * 9.8 m/s^2 * 0.386 m
Multiplying and canceling units:
0.01 kg * v1^2 = 0.02 kg * 9.8 m^2/s^2 * 0.386 m
0.01 kg * v1^2 = 0.076 kg * m^2/s^2
Dividing both sides by 0.01 kg:
v1^2 = 0.076 kg * m^2/s^2 / 0.01 kg
v1^2 = 7.6 m^2/s^2
Taking the square root of both sides:
v1 = sqrt(7.6 m^2/s^2)
v1 ≈ 2.758 m/s
Therefore, the velocity of the bullet just before impact is approximately 2.758 m/s.
To determine the velocity of the bullet just before impact, we can use the law of conservation of momentum.
The law of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, assuming there are no external forces acting on the system.
In this case, the momentum before the collision is equal to the momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v):
p = mv
Let's denote the initial velocity of the bullet as vb and the final velocity after the collision as v'.
Since the bullet collides with the pendulum, the total momentum before the collision is equal to the momentum of the bullet, which is:
p_initial = m_bullet * v_bullet = 0.02kg * vb
After the collision, the pair swings up to a maximum height, indicating that all the initial kinetic energy of the bullet is converted into potential energy of the pair at the maximum height reached.
To find the velocity of the pendulum just after impact, we can equate the initial kinetic energy of the bullet and the potential energy of the pair at maximum height:
0.5 * m_bullet * vb^2 = m_pair * g * h_max
Where m_pair is the combined mass of the bullet and the pendulum, g is the acceleration due to gravity, and h_max is the maximum height reached.
We can solve this equation to find vb:
vb^2 = (2 * m_pair * g * h_max) / m_bullet
vb = √(2 * m_pair * g * h_max) / m_bullet
Substituting the given values:
m_bullet = 0.02kg
m_pair = 0.02kg + 5.75kg = 5.77kg
g = 9.81m/s²
h_max = 0.386m
We can calculate the velocity of the bullet just before impact:
vb = √(2 * 5.77kg * 9.81m/s² * 0.386m) / 0.02kg
vb ≈ 15.69 m/s
Therefore, the velocity of the bullet just before impact is approximately 15.69 m/s.