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
4.9 ms^{-2} in vertical direction
D.
4.9 ms^{-2} in vertical direction
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 sin^{2}θ
Therefore, the relative vertical acceleration of A with respect to B is
g (sin^{2}60 - sin^{2}30) = g/2 = 4.9 ms^{-2} (in vertical direction)
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.
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 λ_{A} to λ_{B} after the collision is
D.
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}
17.57 kg ms^{−1}
A.
17.6 kg ms^{−1 }
P = mv = 3.513 × 5.00 ≈ 17.6