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Practice each question, then open the answer or worked solution.
A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal.
a) find an expression of range on the plane surface
b) time of flight
c) β at which range will be maximum
a) Time of flight, T =
b) Range down an inclined plane, R =
c) Maximum grange down an inclined plane, Rmax =
A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed v0 and rebounds elastically. Find the distance along the plane where it will hit second time.
When x and y axes are selected as shown in the diagram, the motion of projectile from O to A is
y = 0
uy = vo cos θ
ay = -g cos θ
t = T
With the help of kinematics for y-axis
y = uyt + 1/2 ayt2
T = 2vo cos θ/g cos θ
With the help of kinematics for x-axis
x = uxt + 1/2 axt2
A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45o to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?
Let Vrg be the velocity of the rain drop that appears to the girl.
All the vectors are drawn in reference to the frame from the ground.
Let the velocity of rain be:
Case I
According to the problem, the velocity of girl = vg
Let vrg be the velocity of rain with respect to the girl
The equation which is used for determining a is:
Therefore, a = 5
Case II
Now the velocity of the girl has increased = vg
Following is the equation that is used for determining the value of b at tan 45o.
Therefore, the velocity of rain is:
Speed of rain is:
A river is flowing due east with a speed 3 m/s. A swimmer can swim in still water at a speed of 4 m/s.
a) if swimmer starts swimming due north, what will be his resultant velocity?
b) if he wants to start from point A on south bank and reach opposite point B on north bank,
i) which direction should he swim?
ii) what will be his resultant speed?
c) from two different cases as mentioned in a) and b) above, in which case will he reach opposite bank in shorter time?
Given,
Speed of the river, vr = 3 m/s
Velocity of swimmer, vs = 4 m/s
a) When the swimmer swims due north, the y-component has the velocity 4 m/s and x-component will have 3 m/s. following is the resultant velocity with respect to tan θ
θ = 37oN
b) Following is the resultant speed of the swimmer when he wants to reach the point which is directly opposite ti him and which is with respect to tan θ
θ = tan-1(3/√7) of north
c) Time taken in case a) is d/4 sec
Time taken in case b) is d/√7
As d/4 < d/√7, time taken in case a) is shorter than case b).
A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle θ to the horizontal, find
e) how does θ for maximum range change if u > vo, u = vo and u < vo
f) how does θ in e) compare with that for u = 0?
Initial velocity in x-direction,
ux = u + vo cos θ
Initial velocity in y-direction,
uy = vo sin θ
where θ is the angle of projection
a) The angle of projection with the horizontal with respect to spectator is:
tan θ = uy/ux
b) When the displacement along the y-axis is zero over the time period T
y = 0
uy = vo sin θ
ay = -g
t = T
We know that,
y = uyT + 1/2 ayt2
Solving for T, we get
T = 2 uo sin θ/g
c) Horizontal range is
d) For horizontal range, dR/dθ = 0
Then θmax is:
e) If u = vo,
θmax = π/2
f) If u = 0, θmax = 45o
Then, u = 0 and θ ≥ 45o
Motion in two dimensions, in a plane can be studied by expressing position, velocity, and acceleration as vector in Cartesian coordinates
where
and
are unit vectors along x and y directions, respectively and Ax and Ay are corresponding components of A. Motion can also be studied by expressing vectors in circular polar coordinates as
where
are unit vectors along direction in which r and θ are increasing.
e) find velocity and acceleration in polar vector representation for particle moving along spiral described in d) above
a) The unit vector is given as:
Solving the above equations we get unit vector as:
b)
c) Given,
Solving the above gives:
d) Given,
e) Given,
a = 1 unit
Differentiating the above equation, we get velocity as:
By differentiating the above equation we can find acceleration as:
A man wants to reach from A to the opposite comer of the square C. The sides of the square are 100 m. A central square of 50 m × 50 m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he walk only at a speed of v m/s. What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?
As shown in the diagram, APQC is the path taken by the man through the sand, time taken by him to go from A to C,
= 50√2(1/v + 1)
The shortest path will be ARC from the diagram.
Toutside = AR+RC/1 s
RC = AR = 25√10 s
Tsand < Toutside
V > 0.81 m/s