Problem 7 - Master Resource Book in JEE Main Physics

Question

A car is travelling at a velocity of \(10 \mathrm{kmh}^{-1}\) on a straight road. The driver of the car throws a parcel with a velocity of \(10 \sqrt{2} \mathrm{kmh}^{-1}\) when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with direction of the car (a) \(135^{\circ}\) (b) \(45^{\circ}\) (c) \(\tan ^{-1}\left(\sqrt{2)} 60^{\circ}\right.\) (d) \(\tan \left(\frac{1}{\sqrt{2}}\right)\)

Step-by-Step Solution

Answer

Option (a) \(135^{\circ}\) is the correct answer.

Step 1: Understanding the Motion

The car is moving with a velocity of \(10 \mathrm{kmh}^{-1}\) along a straight path. Concurrently, a parcel is thrown from the car with a speed of \(10 \sqrt{2} \mathrm{kmh}^{-1}\). The throw should be such that the parcel reaches a man standing still on the side of the road.

Step 2: Relative Velocity Concept

In order for the parcel to reach the man, the relative velocity of the parcel with respect to the ground should direct towards the man. This means the relative horizontal component of the parcel's velocity should be zero when viewed from the man's frame.

Step 3: Breaking Down Parcel Velocity

The velocity given, \(10 \sqrt{2} \mathrm{kmh}^{-1}\), needs to be resolved into two perpendicular components: one parallel to the car's velocity and one perpendicular to it. These resolve into \(v \cos \theta\) and \(v \sin \theta\) respectively, where \(v\) is the velocity of the parcel and \(\theta\) is the angle of throw with the direction of the car.

Step 4: Setting Up Equations

To make the horizontal component (velocity in the car's direction) zero relative to the man, it must counter the car's velocity. Therefore, the equation is: \(v \cos \theta = 10 \mathrm{kmh}^{-1}\).

Step 5: Solving for the Angle

Plugging the values into the equation, we get \(10 \sqrt{2} \cos \theta = 10\). Simplifying, this leads to \(\cos \theta = \frac{1}{\sqrt{2}}\), which is characteristic of an angle of \(45^{\circ}\). However, since the horizontal must negate the forward motion relative to the standing man, the negative angle is inherently considered as\(180^{\circ} - 45^{\circ} = 135^{\circ}\) in standard position.

Step 6: Determining Correct Angle Option

The angle \(135^{\circ}\) corresponds to the correct direction of throw to make the velocity vector horizontal component zero so that the man receives the parcel. Thus, the correct direction of the throw is option (a).

Key Concepts

Relative Velocity

When it comes to understanding projectile motion, relative velocity is an essential concept. It's about how one object's velocity appears from the viewpoint of another object. To solve problems like the one with the car and parcel, you need to understand how the movement of one object (the parcel) appears from the frame of reference of another object (the man on the side of the road). The driver and the man see the parcel differently because of their different perspectives on its motion.

In the problem, the car and the parcel are in motion, so the man sees the parcel move relative to the car's velocity.

• Relative velocity involves considering how fast one object appears to be moving from the perspective of another moving object.
• This involves subtracting the velocity of the car from that of the parcel to find its velocity relative to the standing man.

If the horizontal component of the parcel's velocity matches the car's velocity but in the opposite direction, to the observer standing still, it seems as though the parcel is simply being dropped vertically. This makes the horizontal displacement zero, allowing it to hit the target - the man.

Vector Resolution

In physics, vectors play a crucial role because they have both magnitude and direction. When you have a situation like the car throwing a parcel, resolving vectors is a key skill to break up a complex motion into simpler parts.

Vector resolution involves splitting a vector into perpendicular components, usually horizontal (x-direction) and vertical (y-direction). This makes it easier to analyze situations where objects move at an angle.

• In the parcel problem, the parcel's velocity is initially given as a single vector, moving both forward and sideways.
• Breaking it down, the velocity resolves into two components: along the direction of the car, and perpendicular to it.

Using basic trigonometry, we determine these components:

• The horizontal component is found using the equation: \(v \cos \theta\).
• The vertical component uses: \(v \sin \theta\).

This decomposition is important because it lets us calculate exactly how fast and in what direction each piece of the projectile's velocity is moving.

Angle of Projection

The angle at which a projectile is launched, also known as the angle of projection, significantly influences its path. For a projectile to land precisely on a target, careful consideration of this angle is crucial. In our problem, the driver throws the parcel at an angle to ensure it falls where the man is standing. This involves an understanding of constructive and destructive force components at play.

The angle of projection is central because:

• It determines the ratio of vertical and horizontal components of velocity.
• It affects how far the projectile will travel both upward and forward.

For the parcel to reach with zero horizontal velocity relative to the man, the angle of upward throw should counter the car's forward motion. Given the velocity, solving for \(\theta\) using trigonometric relationships reveals the needed angle of throw. By solving that \(\cos \theta = \frac{1}{\sqrt{2}}\), we find the actual angle of projection.

• This implies a throw angle of \(135^\circ\) effectively nullifies the excess forward motion caused by the car, making the parcel's path correct for the man to catch it.

This strategic determination ensures the parcel hits the desired point, showcasing the importance of selecting the right projection angle to match the desired trajectory.