HC
Verma Concepts of Physics Solutions - Part 1, Chapter 3 (Rest and Motion: Kinematics):
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OBJECTIVE –I
1. A motor car is going due north at a speed of 50
km/h. It makes a 90° left turn without changing the speed. The change in the
velocity of the car is about
(a) 50 km/h towards west
(b) 70 km/h towards
south-west
(c) 70 km/h towards
north-west
(d) Zero.
Sol: (b)
Change in velocity, Δ V
→ Δ V = V2
– V1
Or, Δ V = √ [502 + (– 50)2]
Or, Δ V = √ (5000)
Or, Δ V = 70.7 ≈ 70 km/h towards south-west.
2. Figure (3-Q2) shows the displacement-time graph of
a particle moving on the X-axis.
(a) The particle is
continuously going in positive x direction
(b) The particle is at rest
(c) The velocity increases
up to a time t0, and then becomes constant
(d) The particle moves at a
constant velocity up to a time t0, and then stops.
Sol: (d)
Slope of the tangent of a
curve at a point in displacement-time graph is equal to the velocity. Here up
to a time t0, slope of the curve remain constant. After that slope
is zero. So, the particle moves at a constant velocity up to a time t0,
and then stops.
3. A particle has a velocity u towards east at t = 0.
Its acceleration is towards west and is constant. Let XA and XB
be the magnitude of displacements in the first 10 seconds and the next 10
seconds
(a) XA < XB
(b) XA = XB
(c) XA > XB
(d) The information is
insufficient to decide the relation of xA with xB.
Sol: (d)
Data is insufficient
to decide the relation of xA with xB.
4. A person travelling on a straight line moves with a
uniform velocity v1 for some time and with uniform velocity v2
for the next equal time. The average velocity v is given by
(a) v = (v1 + v2)/2
(b) v = √( v1 * v2)
(c) 2/v = (1/v1)
+ (1/v2)
(d) 1/v = (1/v1)
+ (1/v2)
Sol: (a)
Given: t1
= t2 = t.
Total distance
travelled, S = v1t + v2t
Total time taken,
T = t + t = 2t
So, the average
velocity, v = S/T
Or, v = (v1
+ v2)/2.
5. A person travelling on a straight line moves with a
uniform velocity v1 for a distance x and with a uniform velocity v2
for the next equal distance. The average velocity v is given by
(a) v = (v1 + v2)/2
(b) v = √( v1 * v2)
(c) 2/v = (1/v1)
+ (1/v2)
(d) 1/v = (1/v1)
+ (1/v2)
Sol:
Time taken by person for travelling
1st x distance, t1 = x/v1.
Time taken by person for travelling
next x distance, t2 = x/v2.
Total time taken, t = t1
+ t2 = x {(1/v1) + (1/v2)}
Total distance travelled, S
= x + x = 2x
So, the average
velocity, v = S/t
Or, v = 2/ {(1/v1)
+ (1/v2)}
Or, 2/v = (1/v1)
+ (1/v2).
6. A stone is released from an elevator going up with
an acceleration a. The acceleration of the stone after the release is
(a) a upward
(b) (g – a) upward
(b) (g – a) upward
(c) (g – a) downward
(d) g downward.
(d) g downward.
Sol: (d)
After the release
of the stone, only acceleration due to gravity acts downward.
7. 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 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
(a) vA > vB
(b) vA < vB
(c) vA = vB
(d) the relation between vA
and vB depends on height of the building above the ground.Sol: (c)
Let height of the building be h.
Ball A projected upwards with
velocity u falls back to building top with velocity u downwards. It completes
its journey to ground under gravity.
Therefore, (vA)2
= u2 + 2gh ------- (1)
Ball B starts with downwards velocity
u and reaches ground after travelling a vertical distance h.
Therefore, (vB)2
= u2 + 2gh ------- (2)
From equation 1 and 2
We get, vA = vB.
8. In a projectile motion the velocity
(a) is always perpendicular
to the acceleration
(b) is never perpendicular
to the acceleration
(c) is perpendicular to the
acceleration for one instant only
(d) is perpendicular to the
acceleration for two instants.
Sol:
9. Two bullets are fired simultaneously, horizontally
and with different speeds from the same place. Which bullet will hit the ground
first?
(a) the faster one
(b) the slower one
(c) both will reach
simultaneously
(d) depends on the masses.
Sol:
Both will reach
simultaneously, because both have same vertical motion.
Initial velocity and
acceleration in vertical direction are same for both the cases.
10. The range of a projectile fired at an angle of 15°
is 50 m. If it is fired with the same speed at an angle of 45°, its range will
be
(a) 25 m
(b) 37 m
(c) 50 m
(d) 100 m.
(b) 37 m
(c) 50 m
(d) 100 m.
Sol:
Given: for both the cases
initial velocity (u) is same.
We know,
Range, R = u2 sin
2θ/g
For 1st case:
→ 50 = u2 sin
2*15 /g
Or, u2/g = 100
For 2nd case:
→ R = u2 sin (2*45)/g
Or, R = u2/g =
100 m.
11. Two projectiles A and B are projected with angle of
projection 15° for the projectile A and 45° for the projectile B. If RA
and RB be the horizontal range for the two projectiles, then
(a) RA < RB
(a) RA < RB
(b) RA = RB
(c) RA > RB
(d) the information is
insufficient to decide the relation of RA with RB.
Sol:
We have required the value
of initial velocity (u) to decide the relation of RA with RB.
12. A river is flowing from west to east at a speed of
5 metres per minute. A man on the south bank of the river, capable of swimming
at 10 metres per minute in still water, wants to swim across the river in the
shortest time. He should swim in a direction
(a) due north
(b) 30° east of north
(c) 30° north of west
(d) 60° east of north.
Sol:
He should swim in a
direction due north to across the river in the shortest time.
13. In the arrangement shown in figure (3-Q3), the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed
(a) 2u cos θ
(b) u/cos θ
(c) 2u/cos θ
(d) u cos θ.
Sol:
OBJECTIVE –II
1. Consider the motion of the tip of the minute hand
of a clock. In one hour
(a) the displacement is zero
(b) the distance covered is
zero
(c) the average speed is
zero
(d) the average velocity is
zero.
Sol:
Initial and final position
of the tip of the minute hand of a clock is same in one hour. So, the displacement
of the tip of the minute hand is zero.
And, velocity = displacement/time
= zero.
2. A particle moves along the X-axis as
x = u (t – 2s) + a(t – 2 s)2.
(a) the initial velocity of
the particle is u
(b) the acceleration of the
particle is a
(c) the acceleration of the
particle is 2a
(d) at t = 2 s particle is
at the origin.
Sol:
Given: x = u (t
– 2s) + a (t – 2 s)2.
We have, dx/dt = u + 2a (t – 2 s)
And, d2x/dt2 = 2a
So, initial velocity = dx/dt|t = 0 = u –
4a.
Therefore, option (a) is wrong.
Acceleration, a = d2x/dt2 = 2a
Therefore, option (b) is wrong and option (c) is
correct.
Position of particle at t =
2 s is given by
→ x = u (t – 2s) + a(t – 2 s)2
Or, x = u (2 s – 2s) + a(2 s
– 2 s)2
Or, x = 0
Therefore, option (d) is
correct.
3. Pick the correct statements:
(a) Average speed of a
particle in a given time is never less than the magnitude of the average
velocity.
(b) It is possible to have a
situation in which |dv/dt|
≠ 0 but d|v|/dt = 0.
(c) The average velocity of
a particle is zero in a time interval. It is possible that the instantaneous
velocity is never zero in the interval.
(d) The average velocity of
a particle moving on a straight line is zero in a time interval. It is possible
that the instantaneous velocity is never zero in the interval. (Infinite
accelerations are not allowed.)
Sol:
4. An object may have
(a) varying speed without
having varying velocity
(b) varying velocity without
having varying speed
(c) nonzero acceleration
without having varying velocity
(d) nonzero acceleration
without having varying speed.
Sol:
If speed of an object is
changes then velocity also changes. So option (a) is wrong. In case of uniform
circular motion speed remain constant but direction changes. So it velocity
changes. Therefore, option (b) is correct.
Rate of change of velocity is acceleration. So, option
(c) is wrong.
In case of uniform circular
motion speed remain constant but it has an acceleration. So, option (d) is
correct.
5. Mark the correct statements for a particle going on
a straight line:
(a) If the velocity and
acceleration have opposite sign, the object is slowing down.
(b) If the position and
velocity have opposite sign, the particle is moving towards the origin.
(c) If the velocity is zero
at an instant, the acceleration should also be zero at that instant.
(d) If the velocity is zero
for a time interval, the acceleration is zero at any instant within the time
interval.
Sol:
6. The velocity of a particle is zero at t = 0.
(a) The acceleration at t =
0 must be zero.
(b) The acceleration at t =
0 may be zero.
(c) If the acceleration is
zero from t = 0 to t = 10s speed is also zero in this interval.
(d) If the speed is zero
from t = 0 to t = 10s the acceleration is also zero in this interval.
Sol:
7. Mark the correct statements:
(a) The magnitude of the
velocity of a particle is equal to its speed.
(b) The magnitude of average
velocity in an interval is equal to its average speed in that interval.
(c) It is possible to have a
situation in which the speed of a particle is always zero but the average speed
is not zero.
(d) It is possible to have a
situation in which the speed of the particle is never zero but the average
speed in an interval is zero.
Sol:
8. The velocity-time plot for a particle moving on a
straight line is shown in the figure (3-Q4).
(a) The particle has a
constant acceleration.
(b) The particle has never
turned around.
(c) The particle has zero
displacement.
(d) 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.
Sol:
9. Figure (3-Q5) shows the position of a particle
moving on the X-axis as a function of time.
(a) The particle has come to
rest 6 times.
(b) The maximum speed is at
t = 6 s.
(c) The velocity remains
positive for t = 0 to t = 6 s.
(d) The average velocity for
the total period shown is negative.
Sol: (a)
10. The accelerations of a particle as seen from two
frames S1 and S2 have equal magnitude 4 m/s2.
(a) The frames must be at
rest with respect to each other.
(b) The frames may be moving
with respect to each other but neither should be accelerated with respect to
the other.
(c) The acceleration of S2,
with respect to S1 may either be zero or 8 m/s2.
(d) The acceleration of S2
with respect to S1 may be anything between zero and 8 m/s2.
Sol: (d)
Discussion - If you have any Query or Feedback comment below.
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