Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B?
4.9 ms-2 in horizontal direction
9.8 ms-2 in vertical direction
zero
zero
D.
zero
For the motion of block along inclined plane
mg sin θ =ma
a = g sin θ
where a is along the inclined plane.
The vertical component of acceleration is g sin2θ
Therefore, the relative vertical acceleration of A with respect to B is
g (sin260 - sin230) = g/2 = 4.9 ms-2 (in vertical direction)
A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is
D.
A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
0.16 J
1.00 J
0.67 J
0.67 J
C.
0.67 J
m1u1 + m2u2 = (m1 + m2)v
v = 2/3 m/s
A body of mass m = 3.513 kg is moving along the x-axis with a speed of 5.00 ms−1.The magnitude of its momentum is recorded as
17.6 kg ms−1
17.565 kg ms−1
17.56 kg ms−1
17.56 kg ms−1
A.
17.6 kg ms−1
P = mv = 3.513 × 5.00 ≈ 17.6
For a particle in uniform circular motion, the acceleration an at point P (R, θ) on the circle of radius R is (Here θ is measured from the x–axis)
C.