IISER Mathematics - Dynamics
Exam Duration: 45 Mins Total Questions : 20
The force required to accelerate a train of mass \({ 10 }^{ 6 }\)kg from rest to a velocity of 6m/second in 2 minutes, is
- (a)
\(4\times { 10 }^{ 4 }N\)
- (b)
\(5\times { 10 }^{ 4 }N\)
- (c)
\(6\times { 10 }^{ 4 }N\)
- (d)
None of these
Here u = 0, v = 6 m/sec, t = 120 seconds
v=u+ft⇒6=0+f(20)
⇒f=\(\frac{1}{20}\) m/sec2
P=mf=106\(\times\)\(\frac{1}{20}\)=5\(\times\)104N
A bullet of mass 20 grams fired into a wall with a velocity of 10m/sec loses all its velocity in one second. Then distance covered by the bullet in the wall is
- (a)
3 m
- (b)
4 m
- (c)
5 m
- (d)
6 m
Here v=0, u=10 m/sec
v=u+ft
0=10+f⇒f=-10m/sec2
v2=u2+2fs
0=102=2\(\times\)10\(\times\)s⇒s=5m
A man of weight 60 kg jumps off a railway train running on horizontal rails at 20 km/hr with a packet of 10 kg in his hand. The thurst of the packet on his hand, is
- (a)
0
- (b)
50 kg wt
- (c)
10 kg wt
- (d)
70 kg wt
A particle is projected under gravity (g=9.81 m/\({ sec }^{ 2 }\)) with a velocity of 29.43 m/sec at an elevation of 30°. The times of flight in seconds, to a height of 9.81 m in second, are
- (a)
0.5, 1.5
- (b)
1,2
- (c)
1.5, 2
- (d)
2,3
The angle of projection of a particle when its range on the horizontal plane is \(4\sqrt { 3 } \) times the greatest height attained is
- (a)
15°
- (b)
30°
- (c)
45°
- (d)
60°
Let u be the velocity and α the angle of projection. If R is the range and h the greatest height attained, then
R=2u2\(\frac{sin\alpha cos \alpha}{g}\)
and h=\(\frac{u^2sin^2\alpha}{2g}\)
Now,
\(\frac { { 2u }^{ 2 }sin\alpha cos\alpha }{ g } =4\sqrt { 3 } \left( \frac { { u }^{ 2 }{ sin }^{ 2 }\alpha }{ 2g } \right) \)
ஃ Simplifying, we get
tan α=\(\frac{1}{\sqrt3}\)⇒α=300
From the top of a tower of height 100 m, a ball is projected with a velocity of 10 m/sec. It takes 5 seconds to reach the ground. If g=10 m/\({ sec }^{ 2 }\), then the angle of projection is
- (a)
30°
- (b)
45°
- (c)
60°
- (d)
180°
Let α be the angle of projection and initial velocity be u = 10 m/sec
Then h=(u sin α)t-\(\frac{1}{2}\)gt2
⇒-100=(10 sin α)\(\times\)5-\(\frac{1}{2}\)\(\times\)10\(\times\)52 (∵ t=5 seconds)
⇒ 25=50 sin ga⇒sin α=\(\frac{1}{2}\)⇒α=300
A particle is projected with a velocity of 39.2 m/sec at an angle of 30° to the horizontal. It will move at right angles to the direction of projection, after the time
- (a)
8 sec
- (b)
5 sec
- (c)
6 sec
- (d)
10 sec
Let u and α be velocity and angle of projection and t be the time when particle moves at right angles to the direction of projection. At this time particle is moving at an angle of α- 900 with horizontal. Then
tan(α-900)=\(\frac{4sin\alpha-gt}{4 cos\alpha}\)
or -cotα=tanα-\(\frac{gt}{4 cos\alpha}\)
or -cot 300=tan 300-\(\frac{9.8t}{39.2 cos 30^o}\)
or \(-\sqrt3=\frac{1}{\sqrt3}-\frac{t}{4\times\frac{\sqrt3}{2}}\)
or \(\frac { t }{ 2\sqrt { 3 } } =\frac { 1 }{ \sqrt { 3 } } +\sqrt { 3 } =\frac { 4 }{ \sqrt { 3 } } \)
ஃ t=8 seconds
A cannon ball has a range R on a horizontal plane. If h and \({ h }^{ \prime }\)are the greatest heights in the two possible paths, then R equals
- (a)
\(\sqrt { 4h{ h }^{ \prime } } \)
- (b)
\(\sqrt { 8h{ h }^{ \prime } } \)
- (c)
\(4\sqrt { h{ h }^{ \prime } } \)
- (d)
\(8\sqrt { h{ h }^{ \prime } } \)
Let α,阝 be angles of projections for two paths for a horizontal range R. Then α + 阝 = 900. Let u be the velocity of projection. Then
R=\(\frac{2u^2sin\alpha cos\alpha}{g}\)
h=\(\frac{u^2sin^2\alpha}{2g}\)
h'=\(\frac{u^2sin^2\beta}{2g}\)
hh'=\(\frac { { u }^{ 4 }{ sin }^{ 2 }\alpha { sin }^{ 2 }({ 90 }^{ 0 }-\alpha ) }{ { 4g }^{ 2 } } \) (∵β=900-α)
or 4hh'=\(\frac { { u }^{ 4 }{ sin }^{ 2 }\alpha { cos }^{ 2 }\alpha }{ { g }^{ 2 } } \)
or 16hh'=\(\left( \frac { { 2u }^{ 2 }sin\alpha cos\alpha }{ g } \right) ^{ 2 }={ R }^{ 2 }\)
or R=4\(\sqrt { hh' } \)
If R is the horizontal range of a projectile and h the greatest height, then velocity of projection is
- (a)
\(g\left( h+\frac { { R }^{ 2 } }{ 16h } \right) \)
- (b)
\(2g\left( h+\frac { { R }^{ 2 } }{ 16h } \right) ^{ 1/2 }\)
- (c)
\(2g\left( h+\frac { R }{ 16h } \right) \)
- (d)
NONE OF THESE
If 4 seconds be the time in which a projectile reaches a point in the path and 5 seconds be the time from P till it reaches the horizontal plane through the point of projection, then the height of the point P above the horizontal plane in metres, is
- (a)
40
- (b)
98
- (c)
196
- (d)
NONE OF THESE
Let u be the velocity and α the angle of projection.
Total time of flight = 4 + 5 = 9 seconds
or \(\frac { 2u\quad sin\alpha }{ g } =9\)
or 4sinα=\(\frac{9g}{2}\) ...(i)
Now, if h is the height of the point P, then
h=(u sinα)t-\(\frac{1}{2}\)gt2
=\(\frac{9g}{2}\)\(\times\)4-\(\frac{1}{2}\)g\(\times\)42=18g-8g=10g
=10\(\times\)9.8=98metres
A particle is projected with a given horizontal range;then number of directions in which the particle can be projected, is
- (a)
4
- (b)
3
- (c)
2
- (d)
1
There are always two directions of projections for given horizontal range.
If you want to kick a football to the maximum distance, then the angle at which it should be kicked is
- (a)
45°
- (b)
90°
- (c)
30°
- (d)
60°
Range=\(\frac{u^2sin2\alpha}{g}\)
For maximum range
sin2α=1⇒α=450
A particle is projected vertically upwards and is at a height h after \({ t }_{ 1 }\)seconds and again after \({ t }_{ 2 }\)seconds. Then h equals
- (a)
\(\frac { { t }_{ 1 }{ t }_{ 2 } }{ 2g } \)
- (b)
\(\frac { { t }_{ 1 }{ t }_{ 2 } }{ g } \)
- (c)
\(\frac { 1 }{ 2 } g{ t }_{ 1 }{ t }_{ 2 }\)
- (d)
\(g{ t }_{ 1 }{ t }_{ 2 }\)
If u is the velocity of projection, then
h = ut - \(\frac{1}{2}\) gt2
or gt2- 2ut+2h=0,
which is quadratic in t, with roots as t1 and t2.
ஃ t1t2=\(\frac{2h}{g}\Rightarrow h=\frac{1}{2}g t_1t_2\)
Two stones are projected from the top of a cliff h metres high, with the same speed u so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle \(\theta \) to the horizontal, then \(\theta \) equals
- (a)
\(2g\sqrt { u/h } \)
- (b)
\(2h\sqrt { u/g } \)
- (c)
\(u\sqrt { 2/gh } \)
- (d)
\(\sqrt { 2u/gh } \)
Let \({ R }_{ 1 }\)and \({ R }_{ 2 }\) respectively be the maximum ranges up and down an inclined plane and \({ R }_{ 3 }\)be the maximum range on the horizontal plane. Then \({ R }_{ 1 },\quad { R }_{ 3 },\quad { R }_{ 2 }\) are in
- (a)
A.P.
- (b)
G.P.
- (c)
H.P.
- (d)
Arithmetic geometric progression
Let β be the inclination of the plane with the horizontal. Then
\({ R }_{ 1 }=\frac { { u }^{ 2 } }{ g(1+sin\beta ) } ,{ R }_{ 2 }=\frac { { u }^{ 2 } }{ g(1-sin\beta ) } and\quad { R }_{ 3 }=\frac { { u }^{ 2 } }{ g } \)
Now \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { g }{ { u }^{ 2 } } (1+sin\beta +1-sin\beta )\)
or \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 2g }{ { u }^{ 2 } } =\frac { 2 }{ { R }_{ 3 } } \)
or \(\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } =\frac { 2 }{ { R }_{ 3 } } \)
Thus, R1, R2, R3 are in H.P.
Hence, the correct alternative is (c)
A particle is projected up a smooth inclined plane whose angle of inclination with the horizontal plane is 30°. The particle comes to rest after 2 seconds; then the velocity of projection, is (where g = 9.8/\({ sec }^{ 2 }\))
- (a)
4.9 m/sec
- (b)
9.8 m/sec
- (c)
19.6 m/sec
- (d)
NONE OF THESE
Proceed like question 44, . putting g = 9.8 m/sec2 , t = 2 seconds
The time of flight of a projectile with horizontal range R is T, then angle of projection of the projectile, is given by
- (a)
\({ tan }^{ -1 }\left( \frac { 2g{ T }^{ 2 } }{ R } \right) \)
- (b)
\({ tan }^{ -1 }\left( \frac { g{ T }^{ 2 } }{ R } \right) \)
- (c)
\({ tan }^{ -1 }\left( \frac { gT }{ R } \right) \)
- (d)
\({ tan }^{ -1 }\left( \frac { g{ t }^{ 2 } }{ 2R } \right) \)
Let the velocity of projection be u and angle of projection be α. Then
T=\(\frac{2usin\alpha}{g}\)
R= \(\frac{u^2sin\alpha}{g}\)
\(\frac{T^2}{R}=\frac{4u^2sin^2\alpha}{g^2}\times{g}{u^2(2sin \alpha cos\alpha)}=\frac{2tan\alpha}{g}\)
or \(\alpha=tan^{-1}(\frac{gT^2}{2R})\)
Hence the correct alternative is (d).
Particles are projected from a point O, in the vertical plane with velocity \(\sqrt { 2gh } \), then the locus of the vertices of their paths is
- (a)
an ellipse
- (b)
a parablola
- (c)
a circle
- (d)
NONE OF THESE
A paraticle is projected upward with velocity 10 m/s along a plane inclined to the horizontal at an angle of 30°, the angle of projection being 60°. Then range of the projectile along the plane is
- (a)
5.8 m
- (b)
6.8 m
- (c)
7.8 m
- (d)
7 m
Here u=10m/sec, α=600, β=300
Range along the plane=\(\frac { { 2u }^{ 2 }cos\alpha sin(\alpha -\beta ) }{ g{ cos }^{ 2 }\beta } \)
\(=\frac { 2\times 100\times cos{ 60 }^{ 0 }sin{ 30 }^{ 0 } }{ 9.8\times { cos }^{ 2 }{ 30 }^{ 0 } } \)
\(=\frac { 200\times \frac { 1 }{ 4 } }{ 9.8\times \frac { 3 }{ 4 } } =\frac { 200 }{ 29.4 } =6.8m\)
Hence the correct alternative is (b).
A particle is projected down an inclined plane, the inclination of the plane being \(\beta \), with an initial velocity u and angle of projection is \(\alpha (\alpha >\beta );\) then the time of flight of the projectile, is
- (a)
\(\frac { 2usin(\alpha +\beta ) }{ gcos\beta } \)
- (b)
\(\frac { 2usin(\alpha -\beta ) }{ gcos\beta } \)
- (c)
\(\frac { 2usin\alpha }{ gcos\beta } \)
- (d)
\(\frac { 2usin\left( \alpha -\beta \right) }{ gsin\beta } \)