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STATEMENT-1 : For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant objects appear to be stationary.
STATEMENT-2 : If the observer and the object are moving at velocities r and respectively with reference to a laboratory frame, the velocity of the object with respect to the observer is .
Statement-1 is true. For a moving observer, the near by objects appear to move in the opposite direction at a large speed. This is because the angular speed of the near by object w.r.t observer is large. As the object moves away the angular velocity decreases and therefore its speed seems to be less. The distant object almost remains stationary.
Statement-2 is the concept of relative velocity which states that.
where G is the laboratory frame.
Thus both the statement are true but statement-2 is not the correct explanation of statement-1.
A ball whose kinetic energy is E, is projected at an angle of 45° to the horizontal. The kinetic energy of the ball at the highest point of its flight will be
Let u be the speed with which the ball of mass m is projected. Then the kinetic energy (E) at the point of projection is
When the ball is at the highest point of its flight, the speed of the ball is (Remember that the horizontal component of velocity does not change during a projectile motion).
∴ The kinetic energy at the highest point
From a building two balls A and B are thrown such that A is thrown upwards and B downwards (both vertically with the same speed). If vA and vB are their respective velocities on reaching the ground, then
Ball A is thrown upwards from the building. During its downward journey when it comes back to the point of throw, its speed is equal to the speed of throw. So, for the journey of both the balls from point A to B .
We can apply v2 – u2 = 2gh.
As u, g, h are same for both the balls, vA = vB
A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is
Dividing (i) and (ii) we get
A boy playing on the roof of a 10 m high building throws a ball with a speed of 10m/s at an angle of 30º with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground ?
(d) From the figure it is clear that range is required
The co-ordinates of a moving particle at any time ‘t’are given by x =at 3 and y =bt 3 . The speed of the particle at time ‘t’ is given by
A ball is released from the top of a tower of height h meters.
It takes T seconds to reach the ground. What is the position of the ball at t/3
We know that now for T/3 second, vertical distance moved is given by
∴ position of ball from ground
if then the angle between A and B is
A projectile can have the same range ‘R’ for two angles of projection. If ‘T1’ and ‘T2’ to be time of flights in the two cases, then the product of the two time of flights is directly proportional to.
Th e angle for which the ranges are same is complementary.
Let one angle be θ, then other is 90° –θ
.Hence it is proportional to R.
Which of the following statements is FALSE for a particle moving in a circle with a constant angular speed ?
Only option (b) is false since acceleration vector is always radial (i.e . towards the center) for uniform circular motion.
An automobile travelling with a speed of 60 km/h, can brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be
A ball is thrown from a point with a speed ' v0' at an elevation angle of q. From the same point and at the same instant, a person starts running with a constant speed '
to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?
Yes, the person can catch the ball when horizontal velocity is equal to the horizontal component of ball’s velocity, the motion of ball will be only in vertical direction with respect to person for that,
A car, starting from rest, accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate f/2 to come to rest. If the total distance traversed is 15 S , then
Distance from A to B = S =
On dividing the above two equations, we get
A particle is moving eastwards with a velocity of 5 ms -1 . In 10 seconds the velocity changes to 5 ms-1 northwards.
The average acceleration in this time is
Therefore the directon is North-west.
The relation between time t and distance x is t = ax 2 + bx where a and b are constants. The acceleration is
A particle located at x = 0 at time t = 0, starts moving along with the positive x-direction with a velocity 'v' that varies as v = a√ x . The displacement of the particle varies with time as
A particle is projected at 60o to the horizontal with a kinetic energy K. The kinetic energy at the highest point is
Let u be the velocity with which the particle is thrown and m be the mass of the particle. Then
At the highest point the velocity is u cos 60° (only the horizontal component remains, the vertical component being zero at the top-most point). Therefore kinetic energy at the highest point.
The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is
A body is at rest at x = 0. At t = 0, it starts moving in the positive x-direction with a constant acceleration. At the same instant another body passes through x = 0 moving in the positive x-direction with a constant speed. The position of the first body is given by x1(t) after time ‘t’; and that of the second body by x2(t) after the same time interval. Which of the following graphs correctly describes (x1 – x2) as a function of time ‘t’?
For the body starting from rest
For the body moving with constant speed
Consider a rubber ball freely falling from a height h = 4.9 m onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic.
Then the velocity as a function of time and the height as a function of time will be :
For downward motion v = –gt The velocity of the rubber ball increases in downward direction and we get a straight line between v and t with a negative slope.
The graph between y and t is a parabola with y = h at t = 0. As time increases y decreases.
For upward motion.
The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same.
Here v = u – gt where u is the velocity just after collision.
As t increases, v decreases. We get a straight line between v and t with negative slope.
All these characteristics are represented by graph (b).
A par t icle h as an in i tial vel ocit y of 3iˆ +4ˆj an d an acceleration of 0.4iˆ + 0.3ˆj . Its speed after 10 s is :
A particle is moving with velocity , where k is a constant. The general equation for its path is
A point P moves in counter-clockwise direction on a circular path as shown in the figure. The movement of 'P' is such that it sweeps out a length s = t3 + 5, where s is in metres and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly.
At t = 2s, at = 6 ´ 2= 12 m/s2
∴ Resultant acceleration
For a particle in uniform circular motion, the acceleration ar at a point P(R,θ) on the circle of radius R is (Here θ is measured from the x-axis )
A small particle of mass m is projected at an angle q with the x-axis with an initial velocity v0 in the x-y plane as shown in the figure. At a time the angular momentum of the particle is
where iˆ,ˆj and kˆ are unit vectors along x, y and z-axis respectively.
An object, moving with a speed of 6.25 m/s, is decelerated at a rate given by: dv/dt =-2.5√v
where v is the instantaneous speed. The timetaken by the object, to come to rest, would be:
A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is :
Total area around fountain
A boy can throw a stone up to a maximum height of 10 m.
The maximum horizontal distance that the boy can throw the same stone up to will be :
Two cars of mass m1 and m2 are moving in circles of radii r1 and r2, respectively. Their speeds are such that they make complete circles in the same time t. The ratio of their centripetal acceleration is :
A particle of mass m is at r est at th e or igin at time t = 0. It is subjected to a force F(t) = F0e–bt in the x direction.
Its speed v(t) is depicted by which of the following curves?
A projectile is given an initial velocity of (iˆ + 2ˆj) m/s, where iˆ is along the ground and ˆj is along the vertical. If g = 10 m/s2 , the equation of its trajectory is :
From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle, to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is:
Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first ? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/ s2) (The figures are schematic and not drawn to scale)