302 - Mechanics part 1 Questions Answers
How fast a car can pull 100kg Iron? the energy of car is 10hp
E= ∫Fds
= ∫ma ds
= ∫m( dv / dt) ds
= ∫m dv (ds/dt)= ∫mvdv
= 1/2 mv2
so v =(2E/m) 1/2 now use mass 100kg & given energy after converting in SI units
This answer is given by SARIKA SHARMA and is correct, we made some modifications only
IF 2sin2220 = p+qcosθ, FIND SMALLEST POSITIVE VALUE OF θ, WHEN p=1 AND q = -1.
2sin222.5 = 1- cos45 = 1 + (-1)cos45
so θ = 45
A car moves towards north at a speed of 54 km/h for 1 h. Then it moves eastward with same speed for same duration. The average speed and velocity of car for complete journey is
avg speed = total distance / total time = 54+54/1+1 = 54
avg velocity = total displacement / total time = 54sqrt(2) / 2 = 27sqrt(2)
The acceleration (a) of a body moving along x-axis is given by a=4x3 ,where a is in m/s2 and x is in metre.if at x=0 the velocity of body is 2m/s,then find its velocity at x=4m?
a = 4x3
so dv/dt = 4x3
or dv/dx * dx/dt = 4x3
or dv/dx * v = 4x3
or vdv = 4x3dx
or ∫vdv = 4∫x3dx
or v2/2 = 4x4/4 + C
at x = 0, v = 2
so 4/2 = 0+C
so C = 2
on putting the value of C
we get v2/2 = x4+ C
now put x = 4 and get v
The position x of a particle moving along x-axis at time t is given by equation t=√x +4 , where x is in meter and t is in seconds. Find the position of particle when its velocity is zero.
given that x = (t-4)2
so dx/dt = 2(t-4)
or v = 2t-8
now for v=0, t=4
put t=4 and get x
A body projected upwards is at same height form ground at t=3s and t=7s . Find the maximum height attained by it. ( take g= 10 m/s^2 ).
A body projected upwards is at same height form ground at t=3s and t=7s. It will gain max. height after 5 second
so maximum height = gt2/2 = 10*5*5/2 = 125
another direct method for finding total distance = 5+15+25+35+45 = 125
these are the distances in 1st, 2nd, 3rd and etc sec.
a particle , with an inital velocity v0 in a plane , is subjected to a constant acceleration in the same plane . then in general ,
the path of the particle would be
1) circle
2) ellipse
3) parabola
4) hyperbola
parabola
for understanding this make components of v parallel and perpendicular to a
a train is moving with a velocity of 30m/s. When brakes are applied , it is found that the velocity reduces to 10m/s in 240m.
When the velocity of the train becomes zero , the total distance travelled is??
270m
by v2 = u2-2as
100 = 900-2*a*240
or 800 = 480a
or a = 80/48 = 10/6 = 5/3
so total distance upto rest is u2/2a = 900*3/2*5 = 270m
angle b/w vectors (i^+j^)and(j^+k^) is
1) 60o
2) 90o
3) 300
4) 450
use the formula A.B = |A||B|cosθ
If a particle (strong enough to survive in space) is released in space from earth. Where it will go? What about its motion? Will it revolve around the sun or remain stationary?
it depends on the speed with which it is thrown