Motion in a plane Solved examples

Let us just assume that both the outer walls are equal in height say h and they are at equal distance x from the end points of the parabolic trajectory as can be shown below in the figure.

Now equation of the parabola is
y=bx-cx² ----(1)
y=0 t x=nr=R
where R is the range of the parabola.
Putting these values in equation (1) we get
b=cnr (2)
Now the range RR of the parabola is
R=a+r+2r+a=nr
This gives
a=(n-3)r/2 (3)
The trajectory of the stone passes through the top of the three walls whose coordinates are
(a,h),(a+r,15h/7),(a+3r,h)
Using these co-ordinates in equation 1 we get
h=ab-ca² (4)
15h/7=b(a+r)-c(a+r)² (5)
h=b(a+3r)-c(a+3r)² (6)
After combining (2), (3), (4), (5) and (6) and solving them we get n = 4.

(a) Position vector is given by
r= xi + yj
Therefore
r= (18.0)t i + [4t - 4.90t² ] j
(b)v= dr/dt =d[(18.0)t i + [4t - 4.90t² ] j ]/dt =18 i + (4-9.8t)j
(c) a= dv/dt =-9.8 j
(d) Att=1 sec
r= (18.0)t i + [4t - 4.90t² ] j
r= (18.0)i - .9 j
(e) At t = 100 s,
v=18 i + (4-9.8t)j =18i -5.8 j
(f)At1= 3.00 s,
a = -9.8 j

(a) The speed and acceleration.
- speed is not constant. a = constant = g.
(b) The vertical component of velocity and acceleration.
- a_y ≠ 0 ⇒ vy not constant.
(c) The magnitude of the acceleration and the speed.
- speed is not constant.
(d) The horizontal component of velocity and acceleration.
- ax = 0 ⇒ vx = constant.
(e) The direction of the velocity and acceleration.
- Direction of v changes.
Therefore answer is D.

Initial situation
a₁ = v² /R
Now v=2πR/T
or a₁= 4π² R/T²
On the second track:
a₂= 4π² (4R)/ (.5T)²
=64π² R/T²
Now
a₂ = 16 a₁ =16p
Hence (b) is correct

(d)

(b) as component is $r \cos \theta$ as this is maximum when $\theta$ is zero

(b) As potential and Pressure are scalar quantities

|A+B|=|A|
Squaring both the sides
|A+B|²=|A|²
$(\mathbf{A} + \mathbf{B}).(\mathbf{A} + \mathbf{B})=\mathbf{A}.\mathbf{A}$
$\mathbf{A}.\mathbf{A} + \mathbf{B}.\mathbf{B} + 2 \mathbf{A}.\mathbf{B}=\mathbf{A}.\mathbf{A}$
$\mathbf{B}.\mathbf{B} + 2 \mathbf{A}.\mathbf{B}=0$
$|\mathbf{B}|^2 + 2 \mathbf{A}.\mathbf{B}=0$
Now $|\mathbf{B}|^2$ cannot be negative, hence $\mathbf{A}.\mathbf{B} \leq 0$
Therefore (d) is the correct option

$|\mathbf{A}|= \sqrt {3^2 + 4^2} =5$
Unit vector in the direction of A will be given by
$\widehat{\mathbf{A}} = \frac {\mathbf{A}}{|\mathbf{A}|} = \frac {1}{5} (3 \mathbf{i} + 4 \mathbf{j})$
Also
$|\mathbf{B}|=\sqrt {7^2 +24^2} = 25$
The vector having the same magnitude as B and parallel to A
$=|\mathbf{B}| \widehat{\mathbf{A}}= 25 \times \frac {1}{5} (3 \mathbf{i} + 4 \mathbf{j})=15 \mathbf{i} + 20 \mathbf{j}$
Hence (a) is the correct option

(a),(b) and (c)

Horizontal velocity remains constant through out the motion. Vertical velocity first decreases and reaches zero at highest point and then start increasing and reaches maximum Just before it hits the ground.
(a) So greatest speed happens Just before it hits the ground.
(b) So smallest speed happens when vertical velocity becomes zero.So At the highest point ,smallest speed is achieved
(c) Acceleration is due to acceleration due to gravity and it remains constants through out the motion.a = g = constant.

At highest point, velocity becomes zero and acceleration remains same as acceleration due to gravity

$R= \frac {u^2 \sin 2\theta}{g}$
So when the initial velocity is doubled, range will becomes four times

Both the spheres will reach simultaneously, if air friction is negligible

90°

a. No
b. Yes
c. Yes
d. False. Magnitude is same but direction is opposite

Here u=1000 m/s, $\theta =30^0$
a. Time taken by the bomb to reach maximum height is given by
$t= \frac {u \sin \theta}{g} =\frac {1000 \times .5}{9.8} = 51 s$
b. Time of flight = 2t= 102 sec
c. The maximum height reached is given by
$h=\frac {u^2 \sin^2 \theta}{2g} = 1.27 \times 10^4$ m
d. Horizontal Range is given by
$R= \frac {u^2 \sin 2\theta}{g}=8.83 \times 10^4$ m

|A + B| = |A - B|
|A + B|² = |A - B|²
or
$(\mathbf{A} + \mathbf{B}).(\mathbf{A} + \mathbf{B})=(\mathbf{A} - \mathbf{B}).(\mathbf{A} - \mathbf{B})$
$\mathbf{A}.\mathbf{A} + \mathbf{B}.\mathbf{B} + 2 \mathbf{A}.\mathbf{B}=\mathbf{A}.\mathbf{A} + \mathbf{B}.\mathbf{B} - 2 \mathbf{A}.\mathbf{B}$
$4 \mathbf{A}.\mathbf{B}=0$
$\mathbf{A}.\mathbf{B}=0$
As the scalar product of A and B is zero,vectors A and B are mutually perpendicular

Let t be the time taken to cover the horizontal distance of 500 m from the gun to the target ,then
$t= \frac {500}{5000} = .1s$
During this time,bullet will fall down vertically due to acceleration due to gravity.
Vertical distance covered
$h= \frac {1}{2} gt^2 = .005$ m = 5cm.
Hence the gun should be aimed at 5cm above the target

Let u be the initial velocity abd t be the time to hit the bird
Horizontal distance covered =$u \cos \theta \times t = \frac {ut}{\sqrt {2}}$
Now
$\frac {ut}{\sqrt {2}}= 59.2$ -(1)
Vertical distance covered in t sec
$h= u \sin \theta t - \frac {1}{2}gt^2 = \frac {ut}{\sqrt {2}}- \frac {1}{2} \times 9.8 \times t^2$
Now
$\frac {ut}{\sqrt {2}}- \frac {1}{2} \times 9.8 \times t^2=39.6$ -(2)
From (1) and (2)
$39.6 = 59.2 - \frac {1}{2} \times 9.8 \times t^2$
t= 4 sec
Substituting this value in (1),we get
$u=29.6 \sqrt {2}$ m/s

a. Anti-parallel
b. Parallel
c. both at an angle of 120°

a -> iii
b -> iv
c -> ii
d -> i