Question

A particle moves with a velocity v in a horizontal circular path. The change in its velocity for covering 60° will be -

(a) v√2

(b) v/√2

(c) v√3

(d) v​

Answer 1

Explanation:

hope u will find it helpful

Answer 2

Answer:

• The change in its velocity is v m/s.

Given:

1. Velocity of particle = v.
2. Angle (∅) = 60°

Explanation:

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#Refer the Attachment for figure.

Here there are two velocities " v " and " v' " as shown.

But we know, in circular motion the direction of velocity changes but the magnitude of the velocity remains constant.

Therefore,

|v| = |v'| = v.

Now, in the figure we have constructed a triangle from vectors " v " " v' " and Δv represents change in velocity.

Now, from Triangle law of Vector Addition we can say that,

⇒ Δv + v = v'

Rearranging,

⇒ Δv = v' - v

Applying law of Vector Subtraction,

⊕ R = √P² + Q² - 2 PQ cos∅

Where,

• R is Resultant = Δv
• P is one vector = v'
• Q is one vector = v
• ∅ is angle = 60°

Substituting the values,

⇒ Δv = √(v')² + (v)² - 2 x (v') x (v) x cos 60°

We know, |v| = |v'| = v, Substituting,

⇒ Δv = √(v)² + (v)² - 2 x (v) x (v) x cos 60°

⇒ Δv = √v² + v² - 2 v² x cos 60°

⇒  Δv = √2 v² - 2 v² x cos 60°

∵ cos 60 = 1/2

⇒  Δv = √2 v² - 2 v² x (1/2)

⇒  Δv = √2 v² - v²

⇒  Δv = √v²

⇒  Δv = v.

∴ The change in its velocity is v m/s (Option - d)

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Additional Formulas:

1. |a| = √(a_x)² = (a_y)²
2. tanα = Q sin∅ / P + Q cos∅ (Direction of the Resultant)
3. R = √P² + Q² + 2 PQ cos∅ (Vector Addition)

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