A bullet of mass M is fired with a velocity 50 m/s at an angle theta with the horizontal. At the hig

Q: Why is the velocity of the bullet purely horizontal at the highest point?

Gravity only acts downwards, so it only reduces the vertical component of velocity. At the peak, the vertical velocity becomes zero, leaving only the constant horizontal component, $v_x = v_0 \cos\theta$.

Q: Why do we use Conservation of Linear Momentum and not Kinetic Energy during the collision?

The bullet embeds itself into the bob, making it a perfectly inelastic collision. In such collisions, kinetic energy is lost to heat and deformation, but linear momentum is always conserved because there are no external horizontal forces acting on the system during the brief impact.

Q: How do we know the string slacks at $120^\circ$?

The problem states the string "moves through an angle of $120^\circ$". In vertical circular motion, if a mass doesn't complete the circle and the string isn't rigid, it slacks when the tension drops to zero. This happens when the radial component of gravity exactly equals the required centripetal force.

Q: How did you calculate the height gained by the pendulum?

The height $h$ in a pendulum swing is given by $l - l \cos\alpha$, where $\alpha$ is the angle from the lowest point. Here, $\alpha = 120^\circ$. Since $\cos 120^\circ = -0.5$, the height is $l - l(-0.5) = 1.5l$.

Q: Why are the coordinates of the bob $(R/2, H)$?

The problem states the bob is at the "highest point of its trajectory". For a projectile fired from the origin over level ground, the highest point is reached exactly halfway through its horizontal range ($R/2$), and its vertical position is the maximum height ($H$).

Question

A bullet of mass $M$ is fired with a velocity $50 	ext{ m/s}$ at an angle $	heta$ with the horizontal. At the highest point of its trajectory, it collides head-on with a bob of mass $3M$ suspended by a massless string of length $10/3 	ext{ m}$ and gets embedded in the bob. After the collision, the string moves through an angle of $120^\circ$. Find
(a) the angle $	heta$,
(b) the vertical and horizontal coordinates of the initial position of the bob with respect to the point of firing of the bullet.
(Take $g = 10 	ext{ m/s}^2$)

Sigma Percentile · Accepted Answer

The problem of the ballistic pendulum at the peak of a projectile's trajectory is a beautiful symphony of three distinct physical phenomena: projectile motion, inelastic collisions, and vertical circular motion. Let's break down this elegant problem step by step.

### The Projectile's Peak

Imagine a bullet of mass $M$ fired from the ground with an initial velocity of $50 	ext{ m/s}$ at an angle $	heta$. As it travels along its parabolic path, gravity continuously decelerates its vertical motion.

When the bullet reaches the absolute highest point of its trajectory, its vertical velocity becomes exactly zero. At this fleeting moment, the bullet is moving purely horizontally. Its speed is simply the horizontal component of its initial velocity, which remains constant throughout the flight:

$$v_x = 50 \cos	heta$$

### The Inelastic Collision

Hanging exactly at this peak is a stationary pendulum bob of mass $3M$. The bullet slams into the bob and embeds itself. Because the bullet and the bob stick together, this is a perfectly inelastic collision.

During the brief moment of impact, there are no external horizontal forces acting on the system. Therefore, we can apply the Conservation of Linear Momentum. The initial momentum of the bullet is transferred to the newly formed combined mass of $4M$. Let's call the velocity of this combined mass immediately after the collision $u$:

$$M (50 \cos	heta) = (M + 3M) u$$

Solving for $u$, we get the initial velocity of the pendulum swing:

$$u = 12.5 \cos	heta$$

### The Pendulum Swing and the Slacking String

Now, the combined mass of $4M$ begins to swing upwards like a pendulum. The problem states that the string moves through an angle of $120^\circ$ from its initial vertical position.

This specific angle is a massive clue. In vertical circular motion, if a string is not rigid, it will slacken when the tension drops to zero. The fact that it reaches exactly $120^\circ$ implies that the tension $T$ becomes zero at this precise point.

At $120^\circ$ from the bottom, the string makes an angle of $60^\circ$ with the upward vertical. Let's analyze the forces at this point. The only force providing the necessary centripetal acceleration to keep the mass in circular motion is the component of gravity acting along the string.

Equating the radial component of gravity to the centripetal force, we get:

$$4Mg \cos 60^\circ = \frac{4M v^2}{l}$$

Given that the length of the string $l = 10/3 	ext{ m}$ and $g = 10 	ext{ m/s}^2$, we can solve for the square of the velocity $v$ at this slacking point:

$$v^2 = g l \cos 60^\circ = 10 	imes \frac{10}{3} 	imes \frac{1}{2} = \frac{50}{3} 	ext{ m}^2/	ext{s}^2$$

### The Work-Energy Theorem

To connect the velocity at the bottom of the swing ($u$) to the velocity at the top ($v$), we use the Work-Energy Theorem, or the Conservation of Mechanical Energy.

As the pendulum swings from the bottom to the $120^\circ$ mark, it gains height. The height $h$ gained by a pendulum swinging through an angle $\alpha$ is given by $h = l - l \cos\alpha$. For $\alpha = 120^\circ$:

$$h = l - l \cos 120^\circ = l \left(1 - \left(-\frac{1}{2}ight)ight) = 1.5 l$$

Substituting the length $l = 10/3 	ext{ m}$, the height gained is exactly $5 	ext{ m}$. Now, we equate the total mechanical energy at the bottom to the total mechanical energy at the top:

$$\frac{1}{2} (4M) u^2 = \frac{1}{2} (4M) v^2 + 4Mg h$$

Dividing by $2M$ and rearranging, we find $u^2$:

$$u^2 = v^2 + 2gh = \frac{50}{3} + 2(10)(5) = \frac{350}{3}$$

### Synthesizing the Angle

We now have two different expressions for $u^2$. One comes from the momentum of the collision, and the other from the energy of the swing. By equating them, we can solve for the firing angle $	heta$.

From momentum, we had $u = 12.5 \cos	heta$, which means $u^2 = 156.25 \cos^2	heta = \frac{625}{4} \cos^2	heta$. Equating this to our energy result:

$$\frac{625}{4} \cos^2	heta = \frac{350}{3}$$

Solving for $\cos^2	heta$:

$$\cos^2	heta = \frac{1400}{1875} = \frac{56}{75} \approx 0.7467$$

Taking the square root gives $\cos	heta \approx 0.864$, which perfectly corresponds to an angle of:

$$	heta \approx 30^\circ$$

### Finding the Coordinates

For the final part of the puzzle, we need to find the initial coordinates of the bob. We know the bob is located at the highest point of the bullet's parabolic trajectory.

The horizontal coordinate $x$ is exactly half of the projectile's total range ($R/2$):

$$x = \frac{R}{2} = \frac{v_0^2 \sin 2	heta}{2g}$$

Substituting $v_0 = 50 	ext{ m/s}$ and $	heta = 30^\circ$:

$$x = \frac{50^2 \sin 60^\circ}{20} = \frac{2500 	imes \frac{\sqrt{3}}{2}}{20} = 62.5 \sqrt{3} \approx 108.25 	ext{ m}$$

The vertical coordinate $y$ is the maximum height ($H$) of the projectile:

$$y = H = \frac{v_0^2 \sin^2	heta}{2g}$$

Substituting our values:

$$y = \frac{50^2 \sin^2 30^\circ}{20} = \frac{2500 	imes 0.25}{20} = 31.25 	ext{ m}$$

Thus, the exact coordinates of the bob are **$(108.25 	ext{ m}, 31.25 	ext{ m})$**. A brilliant conclusion to a multi-layered physics masterpiece!

Visualized Solution

Highest Point Velocity: $v_{x} = 50 cos θ$

Inelastic Collision: $p_{i} = p_{f}$

The Pendulum Swing to $12 0^{\circ}$

Dynamics at Slacking Point: $T = 0$

Work-Energy Theorem: $E_{i} = E_{f}$

Calculating Initial Swing Velocity $u$

Solving for Firing Angle $θ$

Horizontal Coordinate: $x = \frac{R}{2}$

Vertical Coordinate: $y = H$

The Sigma Insight: Vertical Circular Motion

The Projectile's Peak

The Inelastic Collision

The Pendulum Swing and the Slacking String

The Work-Energy Theorem

Synthesizing the Angle

Finding the Coordinates

Similar Questions

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $v$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $θ$ at which the speed of the bob is half of that at $A$ , satisfies

A pendulum bob has a speed of $3 m/s$ at its lowest position. The pendulum is $50 cm$ long. The speed of bob when the length makes an angle of $6 0^{\circ}$ to the vertical will be ....... $m/s$ . (Take, $g = 10 m/s^{2}$ )

A small bob tied at one end of a thin string of length 1m is describing a vertical circle, so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is ……… m/s. (Take, $g = 10 m/s^{2}$ )

A small block slides down from the top of hemisphere of radius $R = 3$ m as shown in the figure. The height $h$ at which the block will lose contact with the surface of the sphere is ………… m. (Assume there is no friction between the block and the hemisphere)

Visualized Solution

Highest Point Velocity: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } vx​=50cosθ

Inelastic Collision: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } pi​=pf​

The Pendulum Swing to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 120∘

Dynamics at Slacking Point: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=0

Work-Energy Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ei​=Ef​

Calculating Initial Swing Velocity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u

Solving for Firing Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Horizontal Coordinate: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2R​

Vertical Coordinate: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y=H