Question 1

A projectile is projected with velocity of 40 ~m / s at an angle with the horizontal. If R be the horizontal range covered by the projectile and after t seconds, its inclination with horizontal become

Accepted Answer

At maximum height, inclination with horizontal becomes zero. Time (taken by) projectile to reach maximum height, aligned t & = u g \ u & = g t (i) aligned Since, R=u (2 t) aligned & = R 2 u t \ & = R 2 g t t \ & = R 2 g t^2 (Using(i)) aligned = R 2 g t^2 = R 2 10 t^2 = R 20 t^2

Question 2

A rigid body rotates about a fixed axis with variable angular velocity equal to - t, at the time t, where , are constants. The angle through which it rotates before it stops is

Accepted Answer

= - t. Comparing with = _0- t Initial angular velocity = Angular'retardation = Angle rotated before it stops is ^2 2 . [using . ^2= _0^2+2 ]

Question 3

The range of the projectile projected at an angle of 15^ with horizontal is 50 m . If the projectile is projected with same velocity at an angle of 45^ with horizontal, then its range will be :

Accepted Answer

Range of projectile R= v^2 2 g ( R (2 )) R_1 R_2 = (2 _1 ) (2 _2 ) = (2 15) (2 45) = 30^ 90^ 50 R _2 = 1 2 R _2=100 ~m

Question 4

A particle of mass m is projected with a velocity ' u ' making an angle of 30^ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle

Accepted Answer

Angular momentum, L=m v H=m u 30 H =m u 30^ u^2 ^2 2 g [ H= u^2 ^2 2 g ] = m u^3 2 g 3 2 ( 1 2 )^2= 3 m u^3 16 g

Question 5

A body is thrown with a velocity of 9.8 ms making an angle of 30^ with the horizontal. It will hit the ground after a time

Accepted Answer

Time of flight = 2 u g = 2 9.8 30^ 9.8 =2 1 2 =1 sec

Question 6

A projectile is projected at 30^ from horizontal with initial velocity 40 ~ms ^ -1 . The velocity of the projectile at t =2 ~s from the start will be: (Given g =10 ~m / s ^2 )

Accepted Answer

Given, Initial velocity of projectile, u=40 ~m / s Angle, =30^ Time of flight T = 2 u g = 2 40 1 10 2 =45 ( g =10 ~m / s ^2 ) It means projectile is at maximum height at t= 2 ~s . At maximum height vertical component of velocity is zero. aligned & Velocity at t =2 ~s = V _ x = u =40 30^ \ & =20 3 ~ms ^ -1 . \ & aligned

Question 7

A projectile is projected with the velocity of (3 i +4 j ) m / s . The horizontal range of the projectile will be

Accepted Answer

Given, v =3 i +4 j v=| v |= 3^2+4^2 v=5 ~m / s From figure, = 4 5 and = 3 5 R= v^2 2 g = v^2 2 g = 5 5 2 4 5 3 5 10 =24 ~m

Question 8

A gun fires two bullets at 60^ and 30^ with horizontal. The bullets strike at some horizontal distance. The ratio of maximum height for the two bullets is in the ratio of

Accepted Answer

The bullets are fired at the same initial speed array l H H ^ = u ^ 2 ^ 2 60^ 2 ~g 2 ~g u ^ 2 ^ 2 30^ = ^ 2 60^ ^ 2 30^ \ = ( 3 / 2)^ 2 (1 / 2)^ 2 = 3 1 array

Motion In Two Dimensions — Previous Year Questions & Solutions