Kinematics Problems (16-06-10)

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Kinematics Problems (16-06-10)

Solutions of these problems are discussed in today's lecture.

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Slide 1 : P. K. BHARTI, I.I.T. KHARAGPUR Kinematics (Conceptual Problems) 16.06.2010 © 2007-10 P K Bharti, IIT Kharagpur

Slide 2 : www.vidyadrishti.com Q1. Figure shows the x – coordinate of a particle as a function of time. Find the signs of vx and ax at t = t1, t = t2 and t = t3. Where to find the solutions of these problems? Go through our lecture on Problem Series XI – Part 3 (Kinematics)

Slide 3 : www.vidyadrishti.com Q2. A motor car is going due north at a speed of 50 km/h. It makes a 900 left turn without changing the speed. The change in velocity of the car is about 50 km/h towards west 70 km/h towards south-west 70 km/h towards north-west zero

Slide 4 : www.vidyadrishti.com Q3. A stone is released from an elevator going up with an acceleration a. the acceleration of the stone after the release is a upward (g – a) upward (g – a) downward g downward

Slide 5 : www.vidyadrishti.com Q4. In a projectile motion the velocity is always perpendicular to the acceleration is never perpendicular to the acceleration may be perpendicular to the acceleration for one instant only is perpendicular to the acceleration for two instants

Slide 6 : www.vidyadrishti.com Q5. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first? The faster one The slower one Both will reach simultaneously Depends on the masses

Slide 7 : www.vidyadrishti.com Q6. A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and the ball B is thrown vertically downward with the same speed. The ball A hits the ground with a speed vA and the ball B hits the ground with a speed vB. We have vA > vB vA < vB vA = vB The relation between vA and vB depends on the height of the building above the ground.

Slide 8 : www.vidyadrishti.com

Slide 9 : www.vidyadrishti.com Q8. Mark the correct statement/s for a particle going on a straight line If the velocity and acceleration has opposite sign, the object is slowing down. If the position and velocity has opposite sign, the particle is moving towards the origin. If the velocity is zero at an instant, the acceleration should also be zero at that instant. If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Slide 10 : www.vidyadrishti.com Q9. The velocity time plot for a particle moving on a straight line is shown in the figure The particle has a constant acceleration. The particle has never turned around. The particle has zero displacement. The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s .

Slide 11 : www.vidyadrishti.com Q10. A particle moves in the plane xy with a constant acceleration a directed along the negative y-axis. The equation of motion of the particle has the form y = ax – bx2 where a and b are positive constants. Find the velocity of the particle at the origin of the coordinates.

Slide 12 : P. K. BHARTI, I.I.T. KHARAGPUR Q. 11. A projectile is launched with a speed u=25m/s from the floor of a 5m high tunnel as shown. Determine the maximum horizontal range R of the projectile and the corresponding launch angle ?. Hint: For maximum range projectile must pass just touching the ceiling of the tunnel. For this maximum height = 5m ? H = 5m ? = 23.327o (Ans) R = 46.397m (Ans)

Slide 13 : www.vidyadrishti.com Q12. A particle is projected from point O on the ground with velocity 5v5 m/s at an angle a = tan – 1 (0.5). It strikes at a point C on a fixed smooth plane AB having inclination of 370 with horizontal as shown in figure. If the particle does not rebound, calculate Coordinates of point C in reference to coordinate system as shown in the figure. Maximum height from the ground to which the particle rises. Take g = 10 m/s2.

Slide 14 : P. K. BHARTI, I.I.T. KHARAGPUR Kinematics (Problems) 16.06.2010 © 2007-10 P K Bharti, IIT Kharagpur Where to find the solutions of these problems? Go through our lecture on Problem Series XI – Part 3 (Kinematics)

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