# Tips to Solve Physics JEE (Advanced) 2018 Questions – IV

1. Home
2. /
4. /
5. Tips to Solve Physics JEE (Advanced) 2018 Questions – IV

### Let Us YOU Solve Physics JEE (Advanced 2018) Questions:

In continuation from the previous blog, we are guiding students to solve JEE (Advanced) 2018, Physics Paper-2 by their own effort. Let us move to Section 2 next. The answer of each problem is a numerical value, up to the second decimal point. Full marks for correct answer; zero marks for incomplete calculation or no attempt.

Question 7:  A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4 kg is at rest on this surface. An impulse of 1.0 N-s is applied to the block at time t = 0 so that it starts moving along the x-axis with a velocity v(t) = v0 e– t / τ , where v0 is a constant and τ = 4 s. The displacement of the block in metres at t = τ  is  ______________ . Take e-1 = 0.37.

The mass of the block is known; the impulse applied to the block is known. How do you calculate the velocity, v0 , acquired by the block immediately after the impulse (at t = 0)? Subscribe to  www.physicsacademyonline.com, and learn from this lecture: Chapter Name – Mechanics; Category –  Basic; Topic Name – Impulse, Collision, and Centre of Mass; Video Name – Impulse, and Impulse-Momentum Theorem.

Next, express instantaneous velocity, v, in derivative form. You may have already watched: Chapter Name – Mechanics; Category –  Basic; Topic Name – Motion in One and Two Dimensions; Video Name – Graphical Analysis of Rectilinear Motion.

Separate the variables, integrate between limits, and the answer follows. You will require some basic technique of integration.

By the way, can you recognise the motion of the block? It is resisted motion under drag force exerted by the oil layer, and the quantity τ  is called “time constant”. Interested students can follow here: Chapter Name – Mechanics; Category –  Basic; Topic Name – Friction; Video Name – Drag Force.

Question 8:  A ball is projected from the ground at an angle 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in metres, is ______________ .

Let the initial speed of projection be u. With what speed does the ball return to the ground? Subscribe to  www.physicsacademyonline.com, and study this lecture: Chapter Name – Mechanics; Category –  Basic; Topic Name – Motion in One and Two Dimensions; Video Name – Projectile Motion. Alternatively, you can decide from: Chapter Name – Mechanics; Category –  Basic; Topic Name – Work and Energy; Video Name – Principle of Conservation of Mechanical Energy, and Some Applications.

Let the speed of projection after the bounce be v. Given the loss of kinetic energy, get a relation between u and v. Relevant lecture: Chapter Name – Mechanics; Category –  Basic; Topic Name – Work and Energy; Video Name – Concept of Energy and Derivation of Work-Energy Theorem.

What is the formula for maximum height reached by a projectile? The lecture on “Projectile Motion”is sufficient. Hence, take the ratio of maximum heights during the two flights. Some quantity cancels out, some quantities are known, and you get your answer.

Finally, once you are through the calculation, verify from the video solution uploaded below.