DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

JEE: DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

The document DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

Introductory Exercise 3.3

Ques 1: Figure shows the displacement-time graph of a particle moving in a straight line. Find the signs of velocity and acceleration of particle at time t = t1 and t = t2.
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Ans: vt1, at1 and at2 are positive while vt2 is negative
Sol: At t = t1
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
v = tan θ
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Corresponding v-t graph will be
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Acceleration at t = t1 : DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
As α < 90°, a t1 is + ive constant.
Acceleration at t = t2

DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 2: A particle of mass m is released from a certain height h with zero initial velocity. It strikes the ground elastically (direction of its velocity is reversed but magnitude remains the same). Plot the graph between its kinetic energy and time till it returns to its initial position.
Ans:
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Sol: Let the particle strike ground at time t velocity of particle when it touches ground

would be gt. KE of particle will beDC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE i.e., KE ∝ t2. While going up the velocity will get - ive but the KE will remain. KE will reduce to zero at time 2 t when the particle reaches its initial position.
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 3: A ball is dropped from a height of 80 m on a floor. At each collision, the ball loses half of its speed. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor. [Take g = 10 m/s2]
Ans: 
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Sol: Speed of ball (just before making first collision with floor) 
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Time taken to reach ground
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Speed of ball (just after first collision with floor)
= 40/2 = 20 m/s
Time to attain maximum height
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
∴ Time for the return journey to floor = 2 s.
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Corresponding velocity-time will be
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 4: Figure shows the acceleration-time graph of a particle moving along a straight line. After what time the particle acquires its initial velocity?
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Ans: (2 + √3) s
Sol:
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
⇒ h = 2 (t - 2)
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Particle will attain its initial velocity i.e., net increase in velocity of the particle will be zero when,
area under a-t graph = 0
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or 3 - (t - 2)2 = 0
or (t - 2)2 = 3
or  t - 2 = ± √3
or   t = 2 ± √3
At time t = 2 + √3 s

Introductory Exercise 3.4

Ques 1: Two balls A and B are projected vertically upwards with different velocities. What is the relative acceleration between them?
Ans:
zero
Sol: Relative acceleration of A w.r.t. B
αAB = (+ g) - (+ g) = 0

Ques 2: In the above problem what is the shape of the graph between distance between the balls and time before either of the two collide with ground?
Ans: straight line passing through origin
Sol: Velocity of A w.r.t. B = vA - vB
∴ Relative displacement (i.e., distance between A  and B) would be
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or   s = (vA - vB) t
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 3: A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10.0 m/s with respect to the water in a direction perpendicular to the river. 
(a) Find the time taken by the boat to reach the opposite bank. 
(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?
Ans: (a) 40 s
(b) 80 m
Sol: In figure, u = speed of boat
v = speed of river flow

DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Time to cross river
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE

Ques 4: An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. Wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s. 
(a) Find the direction in which the pilot should head the plane to reach the point B. 
(b) Find the time taken by the plane to go from A to B.
Ans: DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
(b) 50 min
Sol: Let C be the point along which pilot should head the plane.
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Apply sine formula in Δ ABC
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
= 2989 s
= 50 min

Ques 5: Two particles A and B start moving simultaneously along the line joining them in the same direction with acceleration of 1 m/s2 and 2 m/s2 and speeds 3 m/s and 1 m/s respectively. Initially A is 10 m behind B. What is the minimum distance between them?
Ans: 8 m
Sol: αA = 1 m/s2 ,   αB = 2 m/s2 
vA = 3 m/s,   vB = 1 m/s
Acceleration of A w.r.t. B = 1 - 2 = - 1 m/s2 
Velocity of A w.r.t. B = 3 - 1 = 2 m/s
Initial displacement of A w.r.t. B = - 10 m
At time relative displacement of A w.r.t. B
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or  s = - 10 + 2t - 0.5t2 
For s to be minimum
DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE
or 2 - (0.5 x 2t) = 0
i.e., t = 2 s
∴ smin = - 10 + (2 x 2) - 0.5 x (2)2 
= - 10 + 4 - 2
= - 8 m
Minimum distance between A and B = 8 m.

The document DC Pandey Solutions: Motion in One Dimension- 3 Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

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