302 - Mechanics part 1 Questions Answers
a stone thrown upward with a speed u from the top of a tower reaches the ground with a velocity 4u. The height of the tower is .........
(4u)2 = (-u)2+ 2gh
solve
The velocity of a body depends on time according to the equation v = (t2/10) + 20 . The body is undergoing -
1. Uniform acceleration
2. Uniform retardation
3. Non-uniform acceleration
dv/dt = 2t/10
so a depends on t
option (3)
When a particle is thrown vertically upwards, its velocity at one third of its maximum height is 10√2 m/s. The maximum height attained by it is......
remaining height = 2h/3
so 02 = (10√2)2 - 2g(2h/3)
solve
A body falling from a vertical height of 10m pierces through a distance of 1m in sand. It faces an average
retardation in sand equal to?
[Answer is in terms of 'g' i.e acceleration due to gravity ]
velocity at the time of impact
v2 = 02 + 2g10
now in sand this velocity becomes 0 in 1 m
so 02 = 20g - 2a(1)
so a = 10g
a body thrown vertically up with initial velocity 52 m/s from the ground passes twice a point at h height above at an interval of 10s. The height h is...........
(g=10m/s2)
from that point to topmost point, time taken = 10/2 = 5 sec
so velocity at that point = 50 m/s
now use
522 = 502- 2gh
solve
A bus starts moving with an acceleration of 2m/s2. A cyclist 96m behind the bus starts simultaneously towards the bus at 20 m/s . After what time cyclist will be able to overtake the bus?
let after time t
then distance covered by cyclist in this time = 20t
and by bus = 2t2 / 2 = t2
so according to the given condition
20t = 96+t2
so t = 8,12
after 8 sec cyclist overtakes bus and after 12 sec bus overtakes cyclist
a boat is moved uniformily by pulling with force p=240n,q=200n .what must be the inclination of the resultent force with p and q to have resultent r=400.
In this problem r is the resultant of p and q, let the angle between r and p is α and r and q is β then
p = r sin β/sin(α+β) (1) and q = r sinα/sin(α+β) (2)
so p/q = sinβ/sinα
and according to given condition p/q = 6/5 so sinβ = 6k and sinα = 5k
now obtain cos function and solve it.
a cyclist is moving due east with a velocity of 10 km/h there is no wind air and rain appear to fall at degree vertival calculate the speed of rain.
Question is not submitted properly, angle is not given
let the angle is θ degree
then v(man) = 10i
and v(rain) = -vj i and -j are the directions for man and rain
so v(rain w.r.t. man) = -vj-10i
so tanθ = 10/v so v = 10 cotθ
θ is measured from vertical
A body is moving on a circle of radius 80m with a speed 20m/s which is decreasing at the rate of 5m/s2 at an istant. What is the angle made by its acceleration with its velocity?
You should write that calculate
What is the angle made by its acceleration with its velocity at the instant for which conditions are given.
solution
centripetal acceleration = 20*20/80 = 5 m/s2
tangential acceleration = 5 m/s2
so angle = tan-1(5/5) = 450
A body is projected with a velocity of 20m/s and at an angle of 30° with the ground. The change in the direction of its velocity cannot be (during its motion).
I think options are necessary for this problem.
but if we analyse we can say that the angle will be between 0 and 120 degree