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Sunday, January 3, 2010

2-Kinematics MCQ with Solutions

2-Kinematics MCQ(with Solutions)

1.) At time t = 0 s, a body has starts from rest and moves with constant acceleration. Which graph best represents how s, the displacement of the body, varies with time t?



Answer: E
The parachutist experience free-fall initially, implying the only force acting on him or her is gravitational force. The acceleration will therefore be constant at 9.81 m s-2. The opening of the parachute causes a significant air resistance and hence, causes a resultant force upwards instead of down. Hence, acceleration is now in the opposite direction. The object slows down and hence, air resistance reduces gradually to arrive at terminal velocity when acceleration is zero.


An object is projected at an angle to the horizontal in a gravitational field and it follows a parabolic path, PQRST. These points are the positions of the object after successive equal time intervals. T being the highest point reached. Assuming negligible air resistance, the displacements PQ, QR, RS and ST.

A are equal.
B decrease at a constant rate.
C have equal horizontal components.
D increase at a constant rate.
E have equal vertical components.

Answer: C

The horizontal speed remains constant and hence, the horizontal components are equal.

10

When a rifle is fired horizontally at a target P on a screen at a range of 25 m, the bullet strikes the screen at a point 5.0 mm below P. The screen is now moved to a distance of 75 m and the rifle again fired horizontally at P in its new position.



Answer: C
Initial velocity is zero since object starts from rest. Displacement-time graph should have zero gradient at time t = 0 s.

2 A particle starts from rest and moves in a straight line. Its motion is represented by the acceleration–time graph shown below. At which point is its velocity maximum?



Answer: B

Area under acceleration-time graph represents change in velocity. The particle experiences an increase in velocity in the negative direction continuously to point B. From point B onwards, the velocity decreases until acceleration is zero again, as acceleration and velocity is now in opposite direction (i.e. acceleration positive, velocity negative). Hence, the velocity decreases. The velocity increases again to point C but, the increase is less than the decrease before that, hence overall, the velocity is still lesser than at B.

3.)
An aeroplane, flying in a straight line at a constant height of 400 m with a speed of 300 m s-1, drops an object. The object takes a time t to reach the ground and travels a horizontal distance d in doing so. Taking g as 10 m s-2 and ignoring air resistance, which one of the following gives the values of t and d?
t d
A 1.3 s 0.40 km
B 8.9 s 2.7 km
C 8.9 s 4.0 km
D 40 s 3.0 km
E 40 s 120 km

Answer: B



4.) A body dropped from a tower is timed to take (3.0±0.3) s to fall to the ground. If the acceleration of free fall is taken as 10 m s-2, the calculated height of the tower should be quoted as

A (15.0±0.2) m
B (15±0.1) m
C (45.0±0.2) m
D (45±5) m
E (45±9) m



5.)
A lunar landing module is descending to the Moon’s surface at a steady velocity of 25.0 m s-1. At a height of 40 m, a small object falls from its landing gear.
Taking the Moon’s gravitational acceleration as 1.60 m s-2, at what speed does the object strikes the Moon?

A 22.3 m s-1 B 27.4 m s-1 C 497 m s-1 D 753 m s-1

Answer: B




Take note as u,s and a are all in the same direction, they should have the same sign.

6.)

A parachutist steps from an aircraft, falls freely for 2 s and then opens his parachute. Which graph best represents how his vertical acceleration a, varies with time t during the first 5 s?




Answer: E
The parachutist experience free-fall initially, implying the only force acting on him or her is gravitational force. The acceleration will therefore be constant at 9.81 m s-2. The opening of the parachute causes a significant air resistance and hence, causes a resultant force upwards instead of down. Hence, acceleration is now in the opposite direction. The object slows down and hence, air resistance reduces gradually to arrive at terminal velocity when acceleration is zero.

7.)
When a rifle is fired horizontally at a target P on a screen at a range of 25 m, the bullet strikes the screen at a point 5.0 mm below P. The screen is now moved to a distance of 75 m and the rifle again fired horizontally at P in its new position.



Assuming that air resistance may be neglected, what is the new distance below P at which the screen would now be struck?

A B 15.0 mm C 20.0 mm D 25.0 mm E 45.0 mm

Answer: E
For horizontal direction,
Speed remains constant. Hence, if time needed to travel 25 m is t, time needed to travel 75 m will be 3t.

2 comments:

bala priya said...

nice but not clear

bala priya said...

nice but not clear

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