Motion in a Plane - 1 - Free MCQ Practice Test with solutions, NEET Physics

Option A: True. The velocity vector of a particle at any point on its path is always tangent to the path at that point, as velocity represents the direction of motion.

Option B: True. For a particle in uniform circular motion, the centripetal acceleration points towards the center. Over one complete cycle, the acceleration vectors cancel out, resulting in an average acceleration of zero (null vector).

Option C: True. The net acceleration in uniform circular motion is centripetal acceleration, which always points towards the center of the circle along the radius.

Option D: Not true. This statement is incorrect because it implies that the net acceleration in all types of circular motion is always radial. While in uniform circular motion, the acceleration is purely radial, in non-uniform circular motion, there is an additional tangential component of acceleration due to the change in speed. Thus, the net acceleration in such cases is not always directed towards the center.

When a ball is thrown upwards, the vertical component of its velocity decreases as it rises because of the acceleration due to gravity (g), which acts downward.

• Gravity opposes the upward motion, reducing the ball's vertical velocity until it becomes zero at the highest point.
• After reaching the highest point, the ball starts descending, and the vertical velocity increases in the downward direction.

Thus, as the ball rises, the vertical component of its velocity decreases.

The magnitude of displacement is the shortest straight-line distance between the initial and final positions of the particle.

• If the particle moves in a straight line without changing direction, the displacement is equal to the path length.
• If the particle follows a curved or non-linear path, the displacement is less than the path length.

Thus, the magnitude of displacement is always either less than or equal to the path length of the particle between two points.

Explanation:The motorcycle’s x- and  y-coordinates at t=0.50 s are

The negative value of y shows that the motorcycle is below its starting point.

The motorcycle’s distance from the origin at t=0.50 s is 

Explanation:

The car is in uniform circular motion because it’s moving at a constant speed along a curve that is a segment of a circle. Hence we know 

This is the minimum turning radius because arad is the maximum centripetal acceleration.

Explanation:The flag is fluttering in the north-east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (vwb) of the wind with respect to the boat.

Hence, the flag will flutter approximately due east.

Explanation:

Speed of the man, v_m = 4 km/h
Width of the river = 1 km

Horizontal Motion: The horizontal distance traveled by the ball after time t is: x = u ⋅ t

Vertical Motion: The vertical displacement after time t under gravity is:

Conditions for Hitting the n-th Step:

1. Horizontal distance x must be equal to n⋅b (the horizontal distance to the n-th step):
   
   From this, the time t is:
   
2. Vertical distance y must be equal to n⋅h (the vertical distance to the n-th step):
   y = n ⋅ h

Expand and simplify:

Option A: False. The total path length is the actual distance traveled by the particle, whereas the magnitude of the displacement vector is the straight-line distance between the initial and final positions. These two are only equal if the particle moves in a straight line without changing direction. Otherwise, the path length is greater than the magnitude of displacement.

Option B: True. The magnitude of a vector is always a scalar (it has only magnitude, no direction).

Option C: True. Three vectors not lying in the same plane cannot form a closed triangle (null vector condition), as they cannot satisfy vector addition due to being in different planes.

Option D: True. The average speed is defined as the total path length divided by the time taken, while average velocity is the magnitude of the displacement divided by time. Since the total path length is always greater than or equal to the magnitude of displacement, the average speed is greater than or equal to the magnitude of average velocity.

Thus, Option A is the false statement.

Particle 1 (Thrown Vertically Downward):

Initial velocity (u) is downward.

Final velocity is affected by gravity. Using the equation of motion:

Particle 2 (Thrown Horizontally):

Horizontal velocity remains constant (u).

Vertical velocity due to free fall can be calculated as:

The resultant velocity (v₂) at the Earth's surface is given by combining the horizontal and vertical velocities using the Pythagorean theorem:

Velocity Ratio:
The magnitude of both velocities is the same:

Thus, the ratio of their velocities is: 1 : 1

The described situation is shown in the given figure.

 Here,

v_c = Velocity of the cyclist
v_r = Velocity of falling rain

In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.

Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Since there is no external horizontal force acting on the system (man + trolley), the total momentum of the system remains conserved.

Let v_m ​ = velocity of the man relative to the ground.

v_t ​ = velocity of the trolley relative to the ground.
Initially, both the man and the trolley are at rest. So:
Initial Momentum = 0

When the man walks with velocity 1m/s relative to the trolley:

Momentum of the man relative to the ground:
p_m ​= 80⋅v_m ​

Momentum of the trolley relative to the ground:
p_t = 320⋅v_t ​

From conservation of momentum:

The man’s velocity relative to the trolley is 1m/s:
= v_m ​− v_t ​= 1
Substitute 


Thus, the velocity of the trolley:

Man's velocity relative to the ground:

Displacement of the man relative to the ground after 4s:

The correct displacement is: 3.2 m

The horizontal velocity remains constant throughout the motion because there is no acceleration in the horizontal direction:
v_x = u cosθ

The vertical velocity changes due to the acceleration due to gravity g. Let the vertical velocity at the point where the angle is α be v_y.

 At the given point, the angle α is related to the velocity components by:
 ​
From this, we find:
v_y = v_x​ tan α = u cos θ tan α

 The speed v of the particle at any point is given by:

Substituting 

1.  ​Factor out (u cos θ)² :
   
2. Use the trigonometric identity 1 + tan² α = sec² α:
   
3. Simplify:
   
   
   

(b) Distance between the hostel and the station = 10 km = Displacement of the car

∴ Average velocity = 

Formula for Arc Length:

The distance traveled by the particle is given by the arc length formula:

Here:

• y = x², so dy/dx = 2x
• The limits are a = 0 and b = 2.

Step-by-Step Solution:

1. Differentiate y = x² :
   dy/dx = 2x
2. Square dx/dy :​
   
3. Substitute into the arc length formula:
   
4. Approximation: The integral  cannot be solved exactly using elementary functions, but its value is slightly greater than ​ because ​ grows faster than  ​ over the interval [0, 2].

Why Option c is Correct:

• The exact value of the integral lies just above  , making Option 3 ( > ) the correct choice.

Explanation:

Initial velocity u = 150 m/s

1. Symmetry of Motion:

• Each particle moves towards the next particle, but due to the symmetry of the hexagon, their motion results in a gradual spiral towards the center of the hexagon.
• At any instant, the direction of motion is along the line joining the current position of the particle to the position of the next particle.

2. Relative Velocity Towards the Center:

• The relative motion of the particles ensures that the effective velocity component toward the center is responsible for reducing the distance between them.
• At any moment, the velocity of a particle can be resolved into two components:
  One along the line joining it to the next particle.
  The other perpendicular to it.

The component of velocity along the line joining the particle to the next one is:
v_effective = vcos 
where θ = 60^∘ (the interior angle of a regular hexagon).
Using cos ⁡60^∘ = 1/2
v_effective ​ = v/2

3. Initial Distance to be Covered:

• Initially, the distance between two adjacent particles is the side length of the hexagon, a.
• Due to the symmetry, the particles spiral inward and meet at the center of the hexagon. This meeting point is equidistant from all the initial positions, and the total distance covered by each particle is effectively along its initial direction.

Time to Meet:

The time it takes for the particles to meet can be calculated by dividing the total initial distance a by the effective velocity v_effective :

The velocity of the body at any point during its motion is composed of:

1. Horizontal Component (v_x): This remains constant throughout the motion because there’s no horizontal acceleration.
   
   Using cos ⁡60^∘ = 1/2
   
2. Vertical Component (v_y): This changes with time due to the acceleration due to gravity (g = 9.8 m/s²). At the point where the angle with the horizontal is 30^∘, the relation between the components of velocity is:
    ​ ​
   Using tan ⁡ 
    ​ ​
   Solve for v_y​ :
   
3. Magnitude of the Velocity (v): The total velocity is the vector sum of v_x and v_y​ :
   
   Substitute
   
   

Final Answer:
The speed of the body at the point where it makes an angle of 30^∘ with the horizontal is: 20 m/s

Uniform Circular Motion:

• In uniform circular motion, an object moves along a circular path with a constant speed.
• The direction of motion (and hence the velocity vector) changes continuously due to centripetal acceleration, even though the magnitude of speed remains constant.

1. Angular Momentum:
   • Angular momentum (L) is given by: L = m ⋅ v ⋅ r
     where m is the mass,v is the speed, and r is the radius of the circular path.
   • Both v and r remain constant, so angular momentum is constant.
2. Speed:
   • By definition of uniform circular motion, speed remains constant.
3. Kinetic Energy:
   • Kinetic energy (KE) is given by:
     
     Since m and v are constant, KE is also constant.
4. Momentum:
   • Momentum  is a vector quantity given by:
     
     While the magnitude of momentum (related to speed) remains constant, the direction of  changes continuously as the object moves along the circular path.
   • Therefore, momentum is not constant in uniform circular motion.

Final Answer: The correct answer is: Momentum

Explanation:

(i) PQ = diameter = displacement for each girl = 2r = 2 x 200 = 400 m

Since, displacement vector does not depend upon the actual path length and it is the shortest distance between initial and final position, so in the case of each girl the displacement is 400 m.

(ii)This is equal to the actual length of the path skated by girl B.

 From the given velocity vector:
v_x = k_y(velocity in the x-direction),
v_y = k_x(velocity in the y-direction).

 From the definitions of v_x and v_y, we have:

 Divide the two equations to eliminate t:

 Rearrange the equation:
ydy = xdx.
Integrate both sides:

where C is the constant of integration.
y² = x² + constant.

 The general equation of the path is: y² = x² + constant.
This matches Option b.

Key Concepts:

1. Initial Velocity Components: The particle's initial velocity u has two components:
   • Horizontal velocity: u_x ​= u cos θ,
   • Vertical velocity: u_y ​= u sin θ.
2. Velocity at Any Instant: At any time t, the horizontal and vertical velocities of the particle are:
   • Horizontal: v_x = u cos θ (remains constant),
   • Vertical: v_y = u sin θ−gt, where g is the acceleration due to gravity.
3. Condition for Perpendicularity: At the instant when the velocity vector  becomes perpendicular to the initial velocity  their dot product is zero:
   
   The dot product is:
   
   Substituting the components:
   
4. Solve for t: Expand the equation:
   
   Use the trigonometric identity  ⁡
   
   Solve for t:
   
5. Vertical Velocity at This Instant: At time , the vertical velocity is:
   
   Substitute
   
   Simplify:
   
6. Horizontal Velocity at This Instant: The horizontal velocity remains constant:
   v_x ​= u cos θ.
7. Resultant Velocity v: Since v_y = 0, the resultant velocity is equal to the horizontal velocity:
   v = v_x ​= u cosθ.

 At the point where the particle's velocity is perpendicular to its initial velocity, the magnitude of the velocity is: v = u cot θ, which matches Option c.