EM Practice questions set with answers for class test-II_2023-24
(Q.2) The acceleration of the particle is defined by the relation a = 25 - 3x2 mm/s2. The particle starts with no initial velocity at the position x = 0. (a) Determine the velocity when x = 2 mm (b) the position when velocity is again zero, (c) position where the velocity is maximum and the corresponding maximum velocity.
(Q.3) Acceleration of a particle moving along a straight line is represented by the relation a = 30 - 4.5x2 m/s2. The particle starts with zero initial velocity at x = 0. Determine (a) The velocity when x = 3 m, (b) the position when the velocity is again zero (c) the position when the velocity is maximum.
Solution:
(Q.6) A stone is dropped from the
top of a tower. When it has fallen a distance of 10m, another stone is dropped from a point 38 m below
the top of the tower. If both the stones reach the ground at the same time calculate (i) the height of the tower
(ii) the velocity of the stones when they reach the ground.
(Q.7)
From (v-t) diagram find (i) distance travelled in 10sec. (ii) total distance
travelled in 50 sec.
(iii) Retardation.
(Q.7) From (v-t) diagram find (i) distance travelled in 10sec. (ii) total distance travelled in 50 sec (iii) Retardation.
(Q.8)
The race car starts from rest and travels along a straight
road until it reaches a speed of
(Q.14)
An aero plane is flying in horizontal direction with a
velocity of 540 km/hr and at height of 2200m. When it is vertically above the
point A on the ground, an object is dropped from it. The object strikes the ground at point B. Calculate the distance AB (ignore air resistance). Also find velocity
at B and time taken to reach B.
(Q.20) Figure shows a collar B which moves up with constant
velocity of 2 m/ s. To the collar is pinned
a rod AB, the end A of which slides freely against a 30° sloping ground. For
this instant, determine the angular velocity
of the rod and velocity of end A of the rod.
i)
In triangle IAB,
using sign rule
1.5 IA IB
-------- = --------- = -------
Sin
600 sin 500 sin700
IA = 1.326 m/s and IB = 1.628 m/s
ii)
Rod AB performs
general plane motion
VB = (IB) x (ωAB)
(ωAB) = VB / (IB)
= 2 / 1.628
= 1.23 rad/sec
VA = (IA) x (ωAB)
=
1.326 x 1.23
=
1.63 m/s
In triangle ICB,
(IC)2 = (IB)2 +
(BC)2 – 2(IB)(CB)cos400
= (1.628)2 + (0.75)2 –
2(1.628)(0.75)cos400
= √(1.342)
= 1.158 m
Vc = IC x (ωAB)
= 1.158 x 1.23
= 1.425 m/s
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