Motion In A Plane

In projectile motion, the horizontal component of velocity remains constant throughout the flight because there is no acceleration acting in the horizontal direction (assuming air resistance is neglected). In contrast, the vertical component of velocity changes due to the acceleration due to gravity. Kinetic energy, potential energy, and linear momentum also change during projectile motion due to the variation in both components of velocity. Thus, the horizontal component of velocity remains invariant.

The correct option is (B) : horizontal component of velocity

Given:

• Car moves east with speed .
• Car moves north with speed .

Step 1: Represent Velocities as Vectors

Let the east direction be the positive -axis and the north direction be the positive -axis.

Velocity of : .

Velocity of : .

Step 2: Find Relative Velocity

The velocity of with respect to is:

Step 3: Calculate Magnitude and Direction

Magnitude of :

Direction (angle west of north):

Thus, the direction is west of north.

Conclusion:

The velocity of car with respect to car is , west of north.

Answer:

Given:

• Initial speed of projectile:
• Launch angle:
• Speed at maximum height:

Step 1: Analyze Velocity Components

At maximum height:

• Vertical velocity component ( ) becomes 0.
• Horizontal velocity component ( ) remains constant throughout the flight.

Thus, the speed at maximum height equals the horizontal component:

Step 2: Relate to Initial Velocity

The horizontal component of the initial velocity is:

From Step 1:

Simplify:

Step 3: Find

Using the identity :

Now, calculate :

Conclusion:

The value of is .

Answer:

Given:

• Vector
• Magnitude of vector ,
• Resultant vector

Step 1: Express Vector Addition

The vector addition can be written as:

Step 2: Determine Components of

Let . Then:

Equate the components:

• For component: ⇒
• For component:

Step 3: Use Magnitude of

Given , we have:

Substitute and :

Conclusion:

The magnitude of is .

Answer:

Given:

• Projectile launched at with initial kinetic energy .

Step 1: Resolve Initial Velocity

At the highest point, the vertical velocity component is zero, and only the horizontal component remains:

Step 2: Relate Initial KE to Final KE

Initial KE: .

KE at highest point: .

Since :

Conclusion:

The kinetic energy at the highest point is 10 J.

Answer:

Given:

• Train configuration: 1 engine + 3 coaches (Coach 1 closest to engine, Coach 3 farthest)
• Constant acceleration of train:
• Mass of each coach:

Step 1: Analyze Forces on Coach 2

For Coach 2 to accelerate at , the net force acting on it must be . This force is provided by:

1. The pull from Coach 1 (let's call this )
2. The backward force from Coach 3 (which resists motion)

Step 2: Consider Coach 3's Resistance

Coach 3 has mass and must also accelerate at , requiring force . This force comes entirely from Coach 2 pushing it forward ( ).

By Newton's 3rd Law, Coach 3 exerts an equal and opposite force on Coach 2: .

Step 3: Apply Newton's 2nd Law to Coach 2

The net force on Coach 2 is:

Substitute :

Conclusion:

The force exerted by the first coach on the second coach is .

Answer:

When the elevator moves downward with constant acceleration, the normal force exerted on the person decreases, resulting in the person experiencing weight loss. This is because the downward acceleration reduces the effective weight felt by the person in the elevator.
Hence, the correct answer is (C).