Motion In A Plane

For calculating angle between two vectors we will use the formula

Here therefore |A|= similarly for |B|= , and for dot product we will follow element-wise product of each term which will result into

Therefore ,and this is true when

We know scalar quantities like the energy of an electron can take a negative value. Also, we have an electric charge which is a scalar quantity and can be negative or positive in nature. Also in some cases where inelastic collision is happening energy is not always conserved. Electric Potential is scalar in nature but varies from one point to another in space.

But, it is always true that scalar quantity has the same value irrespective of from where we observing.

When we resolve the U vector, a and b will lie on a positive x-axis and y-axis respectively, suggesting that they are positive in nature (shown in red colour in above figure)

While resolving the orange vector p lies on the positive x- axis while q lies on the negative y- axis. Therefore, a, b, p, are positive while q is negative

When vectors are resolved they have and multiplied to vector’s magnitude. And we know that is always less than 1, this suggests that vectors’ x-component will less than the magnitude of the vector. Therefore, if a vector is lying on the x-axis, then only it will have maximum value as then . (Maximum value of cosθ)

Given in question:

Horizontal range=50m (Assuming it to be R1)

Angle=15° (Assuming it to be θ₁)

And we have to find for

Angle =45° (Assuming it to be θ₂)

Also velocity is same both cases.

Relation between horizontal range R₁ and θ₁ is

Similarly for range R2 and θ2 it is

Dividing both equations,

We know that impulse is change in momentum per unit time and momentum is vector quantity; therefore impulse is also a vector quantity. The area is taken as vector quantity where calculation of flux is required. Area’s vector direction is normal to the its plane

Here it is given speed which is scalar quantity; therefore there is no sense of direction given. Hence we can’t say displacement in equal time intervals are equal, as we don’t know whether the object is moving in a straight line or uniform circular motion

Average acceleration will be zero only when an object is travelling in a straight line, it could without changing its speed, it is moving in a circular motion, which will have non-zero acceleration

When a particle is moving in a circular path with constant speed its average velocity will be zero (displacement will zero at the starting point)

Therefore, with a surety we can only say for constant speed is that it is covering equal path lengths in equal time intervals

It is given that we need to identify a necessarily true statement.

As there is no information of direction, therefore even though the particle’s speed is constant it can be accelerating. For example in a circular motion.

Acceleration depends on the velocity which in-turn depends on its magnitude and direction; therefore no condition can be applied on acceleration.

It could be that particle is moving in a straight line, therefore with a surety, we can’t say the particle is moving in a circular direction.

But, acceleration will always be in the plane of motion only.

We have to identify false statement from above

It is given that

Therefore taking cross product on both side

Now we know that when vectors are parallel then their cross product is zero

Taking post Cross Product on both side with C

Now this could only zero when B and C are parallel to each other as,

only when that’s when B and C are parallel

Therefore statement A is true.

Now taking previous equation

Taking dot product with C on both side

Now this could be zero on two conditions first is that B and C are parallel but it could be zero without C being parallel to B. As when we will take cross product of B and C, then vector perpendicular to both B and C, say vector K. And by taking dot product of K and C it will also be zero as angle between them will always be 90.Therefore B is false

Now if vector triple product of A and B and C, then vector will always lie on place which will formed by A, B and C. This could be visualized by understanding that will always lie in a single plane forming sides of triangle.

Now,

K will be perpendicular to plane containing A and B.

And taking cross product with C (which is also lying on same plane as that of A and B) will give a vector which is perpendicular to C but will be lying on same plane as that of A,B and C. Therefore statement C is true.

It is given in last option that therefore angle between vector A and vector B is 90 and we know that

, therefore form a triangle with angle between A and B equal to 90, therefore it is right angled triangle. Hence option D is also true.

We have to identity statements which are always true. It is given that , it could be true in two conditions that is either , which means option (a) and (b) might true.

For forming a single condition we will multiply them, as either one of them is true it will uphold the necessary condition

We know (from previous equations)

Therefore their magnitude’s product will also be zero.

(This will always be true)

Therefore,

(Equality is true for B=0)

Above condition is always true

Energy is a scalar quantity it doesn’t depend on an angle of projection; it depends only on the speed of projection, which is same for both cases.

It is given that height reached by the first particle is greater than that of second and we know that

Therefore θ₁> θ₂, Now for Time we know that it is directly proportional to Angle therefore, T₁>T₂.

For Range we can’t be sure as for complementary angle they can be equal also.

Energy is always conserved (unless it is an inelastic collision), therefore total energy at point A and P will always be equal as no energy is being lost due to friction.

But, Kinetic Energy will not always remain same, as it depends on the velocity of the particle. When Object is at Point P, its velocity will be zero (Highest point) but the velocity at B will not be equal to zero.

Height at P will be less than Height at A as the object at the highest point has zero velocity that means it is devoid of any kinetic energy, and has only potential energy. Whereas at point A, the object has partly kinetic and partly potential energy (Energy conservation law)

Time Taken from A to B will be more compared to B to P, as the distance from A to B is more compared to B to P.( Height A is more compared to Height P)

When a particle’s acceleration is time-dependent, i.e. acceleration is non-uniform in nature then above two equations aren’t correct

It is given that a particle is performing uniform motion, suggesting that particle’s speed is constant. Also from the figure, we can see that the particle’s velocity vector is always perpendicular to the radius vector. But since, particle’s direction is always changing; therefore, it is accelerating even though speed is constant. Angular momentum’s direction will always be perpendicular to the plane of the circular path.

It is given that , it could be true when |A|=0 or |B|=0 or both are zero.

The given statement can also be true if both conditions are applied together that is

or

Therefore we can write,

This gives the always true condition that is when A is perpendicular to B.

It is given that cyclist maintains its constant speed =10m/s that is the magnitude of velocity is

And we acceleration’s magnitude is given by

We know the acceleration will directed from R to O since this is a centripetal acceleration it will always be towards inside.

Acceleration is only due to gravity; therefore, it will always be pointing toward ground whereas velocity will be directed tangent to path taken by particle.

Greatest speed will be experienced when ball will just reaching the bottom ground, as at time all the potential energy will be completely converted to kinetic energy hence maximum speed will be experienced

Smallest speed will be at highest point, as only single component will be present at time (horizontal), therefore at that time vector sum of velocity will be lowest.

Acceleration is only due to gravity; hence it will remain constant throughout (if we are not considering air drag as this will convert it into non-uniform motion)

Acceleration will always point downward, and it will be due to gravitational force.

At the highest point, ball will stop for some infinitesimally time and change its direction. Therefore at that time velocity will be zero.

Let the cross product of

Therefore, K vector will be perpendicular to both B and C.

And when we will take cross product of K with A

It will be perpendicular to both K and A. Therefore, M will lie on a plane containing vector B and C

Let’s assume the boy throws the ball with vertical velocity u_y and the velocity of the car is u_x.

Therefore, the horizontal velocity of the ball is u_x with respect to the boy on the footpath.

Hence, the motion with respect to the boy on the footpath is parabolic like that of a projectile

The resultant velocity of the ball is

Angle of projection

Given:

Horizontal velocity of car

Vertical velocity of car

Velocity of car

Since the velocity of car is equal to the horizontal velocity of the ball, the ball only has vertical velocity with respect to the car.

Hence, the motion of the ball with respect to the car is a straight line.

In the above figure, the dotted line represents the trajectory of projectile with no air resistance and continuous line represents the trajectory with air resistance.

Because of the air resistance, the horizontal velocity of the projectile is decreasing continuously and so is the energy of the particle. This makes the graph fall steeper than the rise.

Given:

Height of the plane = h = 1.5km = 1500m

Horizontal velocity of the plane = u = 720km/hr = 200m/s

Let us calculate the horizontal distance between the target and the point of release of the bomb,

Now, formula for angle of sight is given by,

(a) Given-

Radius of Earth = R = 6400km

Time period of circular motion of object around the earth = T = 1day = 24 hours

At the equator,

For an object performing circular motion acceleration can be found as,

3.38×10^-2m/s²

At latitude θ,

(b) Given-

Orbital radius of Earth around Sun = 1.5×10¹¹ m

Time period of 1 revolution = 1 year =365 days

We know acceleration during circular motion is given by,

= 5.97×10^-3m/s²

Triangular law of addition states that if two vectors form two sides of a triangle taken in the same order then vector addition of the two vectors completes the third side of the triangle.

From this it is clear that,

In figure(i),

So,

Option (c) matches with figure (i)

In figure(ii),

So,

Option (d) matches with figure (ii)

In figure(iii),

So,

Option (b) matches with figure (iii)

In figure(iv),

Option (a) matches with figure (iv)

Option (a) matches with figure (iv)

Option (b) matches with figure (iii)

Option (c) matches with figure (i)

Option (d) matches with figure (ii)

Given:

|A| = 2, |B| = 4.

(a)

So,

So, θ = 90°

Option (a) matches with (ii)

(b)

So,

So, θ = 0°

Option (b) matches with (i)

(c)

So,

So, θ = 60°

Option (c) matches with (iv)

(d)

So,

So, θ = 180°

Option (d) matches with (iii)

Given:

|A| = 2, |B| = 4.

(a)

So,

So, θ = 0°

Option (a) matches with (iv)

(b)

So,

So, θ = 90°

Option (b) matches with (iii)

(c)

So,

So, θ = 30°

Option (c) matches with (i)

(d)

So,

So, θ = 45°

Option (d) matches with (ii)

Given:

Height of the hill = h = 500m

Speed of projection of packets = u = 125m/s

Distance of canon = 800m

Speed of canon = v = 2m/s

For time to be minimum, the packet should just pass the top of the hill and so the vertical component of velocity must be minimum.

So,

Now let’s calculate the horizontal component of velocity.

Now, for time to reach the top of the hill,

Hence, 20 seconds is the time of flight.

Now, let’s find the horizontal distance between the canon and top of the hill.

R = u_xt = 75m/s×10s =750m

Distance travelled by canon = 800m – 750m = 50m

Time taken by the canon =

Speed of the shells = v_o

Horizontal component of velocity of shells = v_x = v₀cosθ

Vertical component of velocity of shells = v_y = v₀sinθ

Maximum range = R =

We know, maximum range is achieved at θ = 45°

Time of flight = t =

Using kinematic equation,

-h = v₀sinθt – gt²

Substituting the value of t,

-h = v₀sinθ –

Here, θ = 45° , hence sinθ = cosθ =

h = +

∴h = ∆x

Let’s consider the X-axis parallel to the inclined surface and Y-axis perpendicular to it. v be the speed of the projectile.

Range=R, x component of velocity = vcosβ

y component of velocity = vsinβ

x component of acceleration = gsinα

y component of acceleration = gcosα

t is the time of flight

Lets first find part b.

(b) Displacement in y direction is 0.

By 2^nd kinematic equation in y-direction,

(a) By 2^nd kinematic equation in x-direction,

Substituting the value of t in the above equation.

(c) In order to find the value of for which range is maximum,

We need to maximize

Now,

For to be maximum, , should be maximum and the maximum value of sin(θ) is 1.

Therefore,

∴

Let’s consider the X-axis parallel to the inclined surface and Y-axis perpendicular to it. v be the speed of the projectile.

Range=R, x component of velocity = vsinθ

y component of velocity = vcosθ

x component of acceleration = gsinθ

y component of acceleration = gcosθ

There is no displacement in the y direction.

By 2^nd kinematic equation in y-direction,

Now let’s calculate the distance along the plane the particle hit = R,

By 2^nd kinematic equation in x-direction,

Substituting the value of t in the above equation.

Let’s assume the velocity of rain to be

Let the speed of the girl be v_girl

Relative velocity of Rain with respect to the girl is given by,

In the first case,

v_girl = 5m/s

=

The rain falls vertically downwards with respect to the girl, so horizontal component of v_rel = 0

∴ a = 5m/s

In the second case,

v_girl = 10m/s

=

= (-5m/s) î + bĵ

The rain falls at an angle of 45° with respect to the girl, so the horizontal and the vertical components of v_rel are equal.

∴ b = -5m/s

∴

Given:

Speed of the river = v_r = 3m/s

Speed of the man = v_m = 4m/s

(a)

The river flows East and the man swims north

So, resultant speed of the man,

Resultant direction of the man,

(b)

If the man wants to reach the opposite point on the bank, then the horizontal component of his resultant velocity must be zero.

So, v_mx = -v_r = -3m/s

Now, direction in which the man should swim can be found by,

(c) Time taken by man to swim can be found by,

In case a,

In case b,

Clearly, T₁ < T₂

So, the man will reach the opposite bank in shorter time in case a.

Given:

The initial horizontal velocity of the ball with respect to the fielder = v_ocosθ

The initial vertical velocity of the ball with respect to the fielder = v_osinθ

The final horizontal velocity of the ball with respect to the spectator = v_x = u + v_ocosθ

The final vertical velocity of the ball with respect to the spectator = v_y = v_osinθ

(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator can be found as,

tanθ =

(b) Let t be the time of flight.

The displacement in y direction is 0.

(c) Let R be the horizontal displacement.

(d) For R to be maximum, should be equal to 0.

(e) When u = v_o,

When, u << v_o

We can assume,

u << v_o, so we can say that

When, u >> v_o

We can assume,

(f) When u = 0,

=45

(a) We know,

(1)

(2)

Multiplying the equation (1) with sin and (2) with cos

Adding the above two equations,

Substituting in equation (1),

(b) In order to show that and are perpendicular, lets show that their dot product is zero.

Therefore, and are perpendicular.

(c)

Differentiating on both sides,

(d)

Writing dimensions on both sides,

(e) We know,

Here, a=1

Differentiating on both sides to find velocity,

Differentiating on both sides to find acceleration,

When the man walks through sand the shortest path will be,

APQC.

Time taken to travel through sand=T₁=

Here

The shortest path outside sand will be ARC

Here,m

m

Time taken to travel outside sand = T₂=

Now, T₁ < T₂

Hence, the smallest value of v is 0.81m/s