1.A virtual image is formed 20.5 cm from a concave mirror having a radius of curvature of 31.5 cm.
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Answer
given,
mass of the ball = 6.3 kg
speed of the ball = 10.4 m/s
angle made with horizontal = 43°
m_a = 1.8 kg v_a = 2.2 m/s
m_b = 1.6 kg v_b = 1.8 m/s
mass of third particle = 6.3 - 1.8 - 1.6
= 2.9 kg
u cos θ = 10.4 x cos 43° = 7.61 m/s
by using conservation momentum along x-axis
6.3 x 7.61 = 1.8 × (-2.2) + 0 + 1.6 × V₃ₓ
V₃ₓ = 32.44 m/s (toward right)
by using conservation momentum along y-axis
0 = 0 + 1.6 x 1.8 + 1.6 × V₃y
V₃y = -1.8 m/s (indicate downward)
velocity of the third particle
v = 32.49 m/s
θ = 3.176° (downward with horizontal)