Motion In One Dimension (L-1)

• kinetic energy of particle is constant in time

Answer explanation

The correct statement is that the quantity \overrightarrow{v} \times \overrightarrow{a} is constant in time.

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1+β)v, this angle changes to 45°. The value of β is close to :

jee main 2020 (06 sep shift 2)

• 0.50

• 0.41

• 0.37

• 0.73

Answer explanation

The angle change is due to the change in the relative velocity of raindrops with the car. By applying trigonometry, the value of beta is calculated to be close to 0.73, which matches the correct answer choice.

The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is :

• 12m

• 11m

Answer explanation

The total distance covered by the body can be calculated by finding the area under the velocity-time graph. The area between 0 to 4.333 seconds is 37/3 m, and the area between 4.333 to 6 seconds is 0. Therefore, the total distance covered in 6 seconds is 37/3 m.

A balloon is moving up in air vertically above a point A on the ground. When it is a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at an angle 45° with respect to the vertical. When the balloon climbs up a further height h₂, it is seen at an angle 60° with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h₂ is (given tan30°=0.5774):

• 1.464d

• 0.732d

• 0.464d

• d

Answer explanation

The correct height h2 is d because the girl moves 2.464d, which is 4 times the original distance d. Since the angle increases from 45° to 60°, the height also increases by a factor of 4, making it d.

A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity] :

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Answer explanation

The correct choice is t = 3.4*sqrt(h/g) which represents the time taken for the food packet to reach the ground when dropped from the helicopter.

A particle of charge q and mass m is subjected to an electric field E=E₀(1−ax²) in the x-direction, where a and E₀ are constants. Initially the particle was at rest at x= 0. Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is :

JEE Main 2020 (04 sep shift 2)

• a

Answer explanation

The correct choice is sqrt(3/a) because at that distance, the kinetic energy of the particle becomes zero due to the given electric field.

A small ball of mass m is thrown upward with velocity u from the ground. The ball experiences a resistive force mkv² where v is it speed. The maximum height attained by the ball is :

JEE Main 2020 (04 sep shift 2)

Answer explanation

The correct choice is: \(\frac{1}{2k}\ln\left(1+\frac{ku^2}{g}\right)\) which represents the maximum height attained by the ball with resistive force.

The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t=0 to t=5 s will be (a)  

JEE Main 2020 (04 sep shift 2)

Answer explanation

The distance traveled by the particle during the time interval t=0 to t=5 s is equal to the area under the speed-time graph, which is the area of the triangle. Therefore, the distance traveled is (1/2) * base * height = (1/2) * 5 * 10 = 25 m.

A tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height h/2. The velocity versus height of the ball during its motion may be represented graphically by: (graphs are drawn schematically and on not to scale)

JEE Main 2020 (04 sep Shift 1)

• 
• 
• 
• 

Starting from the origin at time t = 0, with initial velocity 5 j ms⁻¹, a particle moves in the x−y plane with a constant acceleration of (10 i+4 j)ms⁻². At time t, its coordinates are (20 m, y₀ m). The values of t and y₀ are, respectively:

JEE Main 2020 (04 sep Shift 1)

• 2s and 18m

• 4s and 52m

• 2s and 24m

• 5s and 25m

Answer explanation

The particle reaches x=20m at t=2s. Using y=5t+0.5(4)t^2, y=5(2)+0.5(4)(2)^2=18m. Therefore, the values are 2s and 18m.