RSM79-P1-KM-PH-55
9.
Solved Problems
9.1
Subjective
Problem 1:
A particle moves along X-axis obeying the equation x = t (t – 1) (t – 2) where x is in metre & t in second. Find the (a) initial velocity and acceleration of the particle (b) velocity & acceleration of the particle when its displacement is zero (c) displacement & acceleration of the particle when its velocity is zero.
Solution :
(a) x = t (t –1) (t – 2) x = t3 – 3t2 + 2t dx v = dt = 3t2 – 6t + 2
. . . (i)
. . . (ii)
At time t = 0, v = 2 m/sec. & dv a = dt = 6t – 6 = 6(t –1) . . . (iii) 2 at time t = 0, a = -6m/sec . (b) Displacement of the particle is zero at time t given by x = t (t – 1) (t-2) = 0 t = 0, t = 1 & t = 2 putting the values of t in eq. (ii) & (iii) we obtain, velocity v = +2, -1 & +2 m/sec. respectively. Acceleration a = -6, 0 & +6 m/sec2 respectively. (c) velocity v =0 v =3t2 – 6t + 2 = 0 1 1 t=1+ 3 &t=1- 3 By putting these values of t in eq. (i) we obtain, 2 2 & x = 3 3 3 3 m respectively & the corresponding acceleration can be obtained by putting the values of t in equation (iii) given by a = 23 m/sec2 & - 23 m/s2 respectively Problem 2:
A car moves from P to Q unidirectionally. It moves with velocities of T T T , & c respectively magnitude aV, bV and cV for interval of times a b for the total path. Find the speed of the car averaged over the time of its motion from P to Q.
Solution:
The average speed
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-56
U1t1 U2 t 2 U3 t 3 t1 t 2 t 3
U
Putting the values of
U1 - aV t1 = T/a
t2 = T/b
U3 = cV
U2 = bV
t3 = T/c
T T T U1 = aV, U2 = bV, U3 = cV, t1 = a , t2 = b & t3 = c we obtain
T T T aV. bV. cV. a b c U T T T b c a 3V `U 1 1 1 a b c Since T = Total time of motion and t1 + t2 + t3 = T T T T 1 1 1 T 1 b c a b c a
. . . (a)
. . . (b)
Using (a) and (b) we obtain U 3V . Problem 3:
An elevator of height h is ascending with an acceleration a. After a time t a coin is dropped from the top of the elevator. When will the coin strike the bottom of the elevator?
Solution:
Let after a time t from the instant of release, the coin strikes the floor of the elevator. The coin was moving with the elevator with a velocity, say v. Just after losing contact it begins to ascend with same velocity v (obeying the law of inertia of motion). Since the net displacement S1 of the coin during time t is downward (w.r.t the initial position or point of release) S1 = ½ gt2 – vt (i) The net displacement S2 of the ascending elevator during time t is upwards (w.r.t. the initial position) S2 = vt + ½ at2 (ii) Now, S1 + S2 = h (iii) Putting S1 and S2 from (i) & (ii) respectively in (iii) we obtain 1 1 2 gt vt vt at 2 h 2 2 1 (g a) t 2 h 2
t
2h (g a)
.
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-57
Problem 4:
Shown in the figure is a ring sliding freely along an inextensible string that passes through the smooth pegs (1) & (2) at the same level. One end of the string is fixed. The other end P is pulled with a constant velocity v, find the velocity of the ring.
Solution:
Pythagoras theorem yields 2 x2 + y2 = , Differentiating both
v
p
x 1
y
2
sides w.r.t. ‘t’, we obtain dx dy d 2x 2y 2 dt dt dt Since the ring slides smoothly, it always lie at the mid position maintaining equal distance from the pegs; Therefore, x remains constant dy d dt = vring , setting dt = rate of pulling of the string = v, we find dy v ring v dt y Putting y = sec , we obtain v = v sec . ring
Problem 5:
A particle is projected up from the bottom of an inclined plane of inclination with velocity v0. If it returns to the point of projection after an elastic impact with the plane, find the total time of motion of the particle.
Solution:
To return to the point of projection after one elastic collision the particle must meet the plane at right angle of the particle along x axis. vx vx0 = (g sin )t0 0 v0 cos = (g sin )t0 v 0 cos t = g sin …(i) 0
x y
plane
v0 g sin
g cos
vx = 0 t = t0
vy0 = v0sin
vy = v0sin
t=0
vx0 = v0cos
Motion of the particle along y axis v y v yo (g cos t0 vy + vy0 = ( gcos ) t0
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-58
2v 0 sin v0 sin + (+v0 sin ) = g cos to t0 = g cos v 0 cos 2v 0 sin Equating (i) & (ii) g sin = g cos cot = 2 tan 2 tan
…(ii)
…(iii)
1
2 cos = 1 tan & sin = 1 4 tan Time of flight for to & fro motion of the particle 2v 0 cos T = 2t = g sin (from (i)) 2
0
2v 0 T = 2t0 =
(2 tan )
1 4 tan 2 g sin
4v 0 2 = g 1 3 sin .
Problem 6:
A particle when fired at an angle = 600 along the direction of the breadth of a rectangular building of dimension 9m 8m 4m so as to sweep the edges. Find the range of the projectile.
Solution :
Since the projectile touches A & B, both of these points lie on the path of the projectile. After putting the coordinate of A in the trajectory equation of projectile we obtain,
v0 (x, h)
{(x + 2h), h } B
A
v0
C x
D
x
2h R
gx 2 2 2 y = h = x tan 2v 0 cos v 02 sin 2 g As we know R = x + x + 2h =
. . . .(i)
v 02 sin 2 g R = 2 (x + h) =
. . . . (ii) g(R / 2 h)2 R 2gR h cos2 2 tan sin 2 Using (i) & (ii) h = 2
1 R R h tan h tan R 2 h= 2
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-59
R2 – 4Rh cot 4h2 = 0
4h cot 16h2 cot 2 16h2 2 R= 1 cos sin = 2h cot (/2) R = 2h (cot + cosec ) = 2h putting = 600 & h = 4m R = 2 4 cot 300 = 8 3 m. Problem 7:
Find the radius of a rotating wheel if the linear velocity v 1 of a point on the rim is 2.5 times greater than the linear velocity v 2 of a point 5 cm closer to wheel axle.
Solution:
Let the radius of the disc = r (in cm.) v 2 (r 5) r v1
(r-5)
v1 is 2.5 times greater than v 2. v2 r 5 r 2.5 v 2
v1
v2 r
r = 2.5 r – 12.5 1.5 r = 12.5 12.5 r = 1.5 = 8.33 cm. Problem 8:
5
A bullet is moving horizontally with certain velocity. It pierces two paper discs rotating co-axially with angular speed separated by a distance . If the hole made by the bullet on 2 nd disc is shifted through an angle with respect to that in the first disc, find the velocity of the bullet. (change of velocity in the bullet is neglected.).
Solution:
Let the bullet take a time t to travel from one disc to the other with a velocity v. The distance between the discs = v.t. = vt t v
v
. . . (a) When the bullet is about to hit the second disc at Q, the hole made by the bullet on the first disc at P will rotate through an t v angle during the time = t
P
t=0
t=t
Q
t
. . . (b)
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-60
Using (a) & (b) ω v θ .
t
v = /
Problem 9 :
A particle is revolving with a constant angular acceleration in a circular path of radius r. Find the time when the centripetal acceleration will be numerically equal to the tangential acceleration.
Solution :
v Let the speed of the particle after at t=t time t from starting be v The ar centripetal acceleration 2 v w v 2 r ar r r & the corresponding angular speed = t ar = r(t)2 = r 2 t2 (i) We know that, the tangential acceleration at = r …(ii) Since, ar = at (given) r 2 t2 = r 1 t α .
t=0
Problem10:
A satellite appears in every T = 10 hour over any place on earth’s surface on equational line. Find the period of revolution of the satellite.
Solution:
Let the period of revolution of the satellite = T s and the period of rotation of earth TE The angular speed of earth and the satellite are 2 2 & s E T TE s respectively The angular velocity of the satellite w.r.t. earth SE = s E 2 2 2 Ts TE TSE 1 1 T T s
1 TE
TTE TE T , put T = 10 hrs & T = 24 hrs. E 10 24 10 24 Ts 10 24 hrs & 24 10 hrs. we obtain Ts = 7.05 hrs & 17.14 hrs. Ts
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-61
9.2
Objective
Problem 1:
A car is moving eastwards with velocity 10 m/s. In 20 sec, the velocity changes to 10 m/s northwards. The average acceleration in this time: (A) 1 2 m s-2 towards N-W (C) 1/2 m s-2 towards N-W
Solution :
(B) 1 2 ms-2 towards N-E (D) 1/2 ms-2 towards N
The change in velocity v v 2 v1 v 10 j 10i v has a magnitude 102 and directed due N – W v 10 2 1 | a | m / s2 t 20 2 and directed due N – W. (A)
y N
450 W
v2
v
E
v1 S
Problem 2:
A driver applies the brakes on seeing traffic signal 400 m ahead. At the time of applying the brakes vehicle was moving with 15 ms -1 and retarding with 0.3 ms-2. The distance of vehicle after 1 min from the traffic light: (A) 25 m (B) 375 m (C) 360 m (D) 40 m
Solution :
The maximum distance covered by the vehicle before coming to rest v2 (15 )2 375 m 2 a 2 ( 0 . 3 ) = v 15 50 s The corresponding time = t = a 0.3 . Therefore after 50 seconds, the distance covered by the vehicle = 375 m from the instant of beginning of braking. The distance of the vehicle from the traffic after one minute = (400 – 375) m = 25 m. (A)
Problem 3:
A particle is moving along a circular path of radius 5m and with uniform speed 5 m/s. What will be the average acceleration when the particle completes half revolution? (A) zero (B) 10 m/s2 2 (C) 10 m/s (D) 10/ m/s2
Solution :
The change in velocity when the particle completes half revolution is given by
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
x
RSM79-P1-KM-PH-62
v = 5 m/s – (-5 m/s) = 10 m/s Time taken to complete half revolution r 5 5 t= v v 10 m / s 10 m / s2 t s Average acceleration = .
Problem 4:
(D)
Which of the following graph correctly represents velocity-time relationship for a particle released from rest to fall freely under gravity? (A)
(B) v
v
t
t
(C)
(D) v
v
t
Solution:
t
Releasing of the particle from rest means that v0 =0 at t = 0 and v =gt at any time t. the slope of v/t graph is a constant. v/t graph is a straight line passing through the origin. (A)
Problem 5 :
Which statements can be possible cases in one/two dimensional motion: (A) A body has zero velocity and still be accelerating (B) The velocity of an object reverses direction when acceleration is constant (C) An object be increasing in speed as its acceleration decreases (D) None of these
Solution :
(a) When the body is projected vertically up, at the highest point its speed becomes zero whereas it is accelerating downwards with g = 9.8 m/s2. (b) When a body is projected up, the velocity during ascent reverses its direction during its descent, whereas the acceleration of the g body remains constant, that is . (c) Acceleration means, increasing speed (magnitude of velocity). Therefore, the speed of a particle increases with an acceleration.
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-63
When the acceleration is decreased the speed of the particle goes on decreasing till the acceleration reduces to zero. (A) and (B) Problem 6:
Solution: Problem 7:
A motor boat is to reach at a point 300 upstream on other side of a river flowing with velocity 5 m/s. Velocity of motor boat with respect to water is 5 3 m/sec. The driver should steer the boat an angle: (A) 300 w.r.t. the line of destination from starting point (B) 600 w.r.t.. normal to the bank (C) 1200 w.r.t. stream direction (D) None of these
Solution :
The velocity of motor boat is given as v m v mw v w v m (5 3 cos 30 )i (5 3 sin 30 ) j 5i 5 3 v m 7.5 i j 5i 2 5 3 v m 2.5 i j 2 5 3 2 2.5 1 = tan = tan1(-3) = 1200. (C)
y vmw
vm 300
v
x
Problem 8:
Choose the wrong statement (A) Zero velocity of a particle does not necessarily mean that its acceleration is zero. (B) Zero acceleration of a particle does not necessarily mean that its velocity is zero. (C) If speed of a particle is constant, its acceleration must be zero. (D) none of these
Solution :
If speed, that is the magnitude of velocity is constant, where as the direction of velocity changes, we cannot say that velocity is constant. Therefore the particle has non-zero acceleration (C)
Problem 9:
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion & air resistance, its velocity V varies with the height h above the ground, as
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942
RSM79-P1-KM-PH-64
(A)
(B) v
v d
h
(C)
(D) V d
Solution:
h
d
V d
h
h
The kinematical equations of motion of the ball
2gh
while falling down v = Just before impact, v1 = Just after impact, v2 = Afterwards, v = (A).
, 2gd
,
2g(d / 2)
v 22 2gh
Problem 10:
The angular acceleration of a particle moving along a circular path with uniform speed is (A) uniform but non-zero (B) zero (C) variable (D) such as cannot be predicted from the given information.
Solution :
As angular speed of the particle is constant and hence angular acceleration is zero. (B)
FIITJEE Ltd., ICES House, Sarvapriya Vihar (Near Hauz Khas Bus Term.), New Delhi - 16, Ph : 26515949 , 26865182, 26854102, Fax : 26513942