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### Kinematics Numerical

1. A boy is standing at a distance a1 from the foot of a tower.The boy throws an stone at a angle 45° which just touches the top of the tower and strikes the ground at a distance a2 from the point the boy is standing.Find the height of the tower

2. A ball is thrown upward from a point on the side of a hill which slopes upward uniformly at angle 45°.Intial velocty of the ball is u0 m/s and at an angle 65°(with respect to the horizotal).At what distance up the slope the ball strike and in what time?

3.A cannon on a level plain is aimed at an angle θ above the horizontal.A shell is fired with a muzzle velocity v0 towards a pole which is distance R away.It hits the pole at height H.
a find the timetaken to reach the pole
b. find the value of H in terms of θ,R and v0

4. The displacement of the body x(in meters) varies with time t (in sec) as
x=(-2/3)t2 +16t+2
find following
a. what is the velocty at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d .what will the displacement when it comes to rest
e .How much time it take to come to rest.

5. A man runs at a speed at 4 m/s to overtake a standing bus.When he is 6 m behind the door at t=0,the bus moves forward and continues with constant acceleration of 1.2 m/s2
find the following
a. how long does it take for the man to gain the door
b if in the beginning he is 10m behind the door ,will he running at the same speed ever catch up bus?

Detailed Solutions:
3. The time taken to reach the pole is given
R=v0cosθ t
or t=R/v0cosθ

Now equation of motion of vertical motion
h=v0sinθt-1/2gt2
at t=R/v0cosθ h=H
so
H=(v0sinθ)X(R/v0cosθ ) -(1/2)g( R/v0cosθ)2
or H=Rtanθ-gR2/2v02cos2θ

4.

Given x=(-2/3)t2 +16t+2

Velocity=dx/dt=(-4/3)t+16
so velocity at t=0 =16
and velocity at t=1 =(-4/3)+16=44/3

Acceleration is given as =d2x/dt2=-4/3

so acceleration is time independent and it is constant

Displacement at t=0 can be found by simply substituting the values of t=0 in equation (1)
=2

Now v=(-4/3)t+16
when v=0
then t=4/3 sec

Displacement can be found by substituing the value of t=4/3 in equation (1)
=-32/27+64/3+2
=(-32+576+54)/27=598/27 m

### Solution for thermo conceptual question

Solution of PART 1
1. (a)
There was no restriction for it expansion.So no tensile or compressive force developed.Longitudinal strains happens only when tensile or compressive force developed in the rod.So answer a

2.(c)
3.(d)
4.(c)
5. (a),(b)
6.(b)
7.(a),(c)
Due to thermal expansion,the diameter of the disc as well of the hole will increase.therefore the moment of inertia will increase resulting in a increase in the angular speed.

Solution of PART 2
1 (c)
2. (b)
3. (b)
4.(c)
5. (a),(c)

Solution of PART 3
1. (a)
dU=dQ-dW or dU=dQ as dQ is negative & dW=o
so dU is negative .dU <0 now dU =nCvdT
so nCvdT <0 or dT<0
2.(d)
3.(c) as dW=pdV so if dW> 0 the dV >0
4. (b),(c) as temperature is constant.
5. (a)
6. (a),(b)
7. (a),(b),(d)
8. (c)
9.(a),(c)
10. (a),(b)

Solution of PART 4
1. (d)
2. (c)
3. (b)
4. (a)
5. (d)
6. (d)
7. (a),(b)
8.(d)

More detailed solution for each question will be given later on