a pendulum consists of a wooden bob of mass m and length l. A Bullet of mass m1 is fired towards the pendulum with a speed v1.the bullet emerges out of the bob with a speed v1/3 and the bob just completes motion along a vertical line ,then v1 is?
Let Vf be the bob velocity just after the bullet has gone through, before it has started to swing upwards.
(1/2)Vf^2 = 2 g l
(from conservation of energy)
Vf = sqrt (4 g*l) = 2 sqrt(g*l)
Now apply conservation of momentum to the bullet interaction:
m1*v1 = m1*(v1/3) + m*Vf
(2/3)m1*v1 = 2*m*sqrt(g*l)
v1 = 3(m/m1)*sqrt(g*l)
What do you mean by "completes motion along a vertical line" ? Does the bob go straight up and stop?
yes the bob goes straight up and stops
What if the motion of the bob after collision just complete along a vertical circle??
To solve this problem, we can make use of the principles of conservation of linear momentum and conservation of mechanical energy.
Let's calculate the initial momentum of the bullet before hitting the pendulum. The momentum (p) is defined as the product of the mass (m1) and velocity (v1) of the bullet.
Initial momentum of the bullet = m1 * v1
Next, let's consider the situation after the bullet has passed through the bob of the pendulum. At that moment, the bob is just completing its motion along a vertical line. Let's assume the velocity of the bob is v2.
According to the principle of conservation of linear momentum, the total momentum before and after the bullet hits the bob should be the same. Therefore, we have:
m1 * v1 = m * v2
Now, let's consider the conservation of mechanical energy. Since the bob completes its motion along a vertical line, we can assume that the initial mechanical energy of the system is equal to the final mechanical energy. The initial mechanical energy consists of the potential energy when the bob is at its highest point, given by m * g * l, where g is the acceleration due to gravity and l is the length of the pendulum. The final mechanical energy consists of the kinetic energy of the bob, which is (1/2) * m * v2^2.
Setting the initial and final mechanical energies equal, we get:
m * g * l = (1/2) * m * v2^2
Now, we have two equations:
m1 * v1 = m * v2 [Equation 1]
m * g * l = (1/2) * m * v2^2 [Equation 2]
We can simplify Equation 2 by canceling out the mass (m) from both sides, which gives:
g * l = (1/2) * v2^2
Substituting Equation 1 into Equation 2, we get:
g * l = (1/2) * (m1 * v1 / m)^2
Rearranging this equation to solve for v1 gives us:
v1 = sqrt(2 * g * l * (m / m1))
So, v1 is equal to the square root of 2 times the gravitational acceleration (g) times the length of the pendulum (l) times the mass of the pendulum (m) divided by the mass of the bullet (m1).