A particle is projected from point A with velocity √18 m s⁻¹ at an angle 45° with horizontal as shown. It strikes the plane BC at right angles. Velocity of the particle at the time of collision is V such that V/2 = √x. Then x = ...
Physics · Mixed Chapter
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Question

A particle is projected from point A with velocity √18 m s⁻¹ at an angle 45° with horizontal as shown. It strikes the plane BC at right angles. Velocity of the particle at the time of collision is V such that V/2 = √x. Then x = ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The projectile starts at speed √18 m/s at 45°, so its horizontal velocity is constant: vₓ=√18 cos45°=3 m/s. Step 2: Plane BC is inclined at 60° to the horizontal. Just before impact the velocity is perpendicular to BC and directed below the horizontal, so vᵧ/vₓ=−tan30°=−1/√3. Therefore vᵧ=−√3 m/s. Step 3: Impact speed V=√(vₓ²+vᵧ²)=√(9+3)=√12=2√3 m/s. Step 4: V/2=√3=√x ⇒ x=3.