Question

Two bodies A and B of masses mA = 1 kg and mB = 2 kg are initially at rest at the same height y0 = 80 cm above a horizontal plane. At t = 0 they are both released. A slides along a plane inclined by an angle θ = 45° while B falls vertically. Let tA and tB be the time required for A and B to reach the horizontal plane, vA and vB are the corresponding speed at tA and tB. Which of the following results is correct? (Assume g = 10 m/s2)
A: vA = 1/2 vB and tA = tB = 0.4 s
B: vA = 2 vB and tA = tB = 0.4 s
C: vA = vB = 4 m/s and tA = √2tB
D: vA = vB = 4 m/s and tA = 1/2tB = 0.8 s
E: vA = vB = 4 m/s and tA = 1/√2tB

Answer 1

Question 77978

Two bodies A and B of masses mA = 1 kg and mB = 2 kg are initially at rest at the same height y0 = 80 cm above a horizontal plane. At t = 0 they are both released. A slides along a plane inclined by an angle θ = 45° while B falls vertically. Let tA and tB be the time required for A and B to reach the horizontal plane, vA and vB are the corresponding speed at tA and tB. Which of the following results is correct? (Assume g = 10 m/s2)

A: vA = 1/2 vB and tA = tB = 0.4 s

B: vA = 2 vB and tA = tB = 0.4 s

C: vA = vB = 4 m/s and tA = √2tB

D: vA = vB = 4 m/s and tA = 1/2tB = 0.8 s

E: vA = vB = 4 m/s and tA = 1/√2tB

Solution

For body A $v_{a} = g \times \sin \alpha \times t_{a}$, for body B $v_{b} = g \times t_{b}$, so $\frac{v_{a}}{v_{b}} = \frac{g \times \sin \alpha \times t_{a}}{g \times t_{b}} = \frac{\sin \alpha \times t_{a}}{t_{b}}$. The correct answer is C, because when $v_{a} = v_{b}$, $\sin \alpha \times t_{a} = t_{b}$, so $t_{a} = \frac{t_{b}}{\sin \alpha} = \frac{t_{b}}{\frac{\sqrt{2}}{2}} = \sqrt{2} t_{b}$.

Answer

Result C is correct, all other are incorrect

Answer provided by https://www.AssignmentExpert.com