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if a particle of mass m strikes on ground with angle of incidence 45 and coefficient of restitution is 1V2 find the velocity of reflection and angle of reflection . i know the answer but how the horizontal velocity component i.e. Vx =V'x i.e. before and after strike , where vertical component are not equal,plz explain

bhupinder sidhu , 13 Years ago
Grade 12th Pass
anser 1 Answers
Askiitians Tutor Team

To tackle this problem, let's break it down step by step. When a particle strikes the ground at an angle, we can analyze its motion by separating it into horizontal and vertical components. The coefficient of restitution plays a crucial role in determining how the particle behaves after the collision. Here’s how we can approach this scenario.

Understanding the Components of Motion

When the particle strikes the ground at a 45-degree angle, we can denote its initial velocity as V. This velocity can be broken down into its horizontal and vertical components:

  • Horizontal component: Vx = V * cos(45°) = V / √2
  • Vertical component: Vy = V * sin(45°) = V / √2

Coefficient of Restitution

The coefficient of restitution (e) is defined as the ratio of the relative speed after the collision to the relative speed before the collision, specifically in the direction perpendicular to the surface. In this case, the coefficient of restitution is given as 1/√2.

Mathematically, this can be expressed as:

e = (V'y) / (Vy)

Where V'y is the vertical component of the velocity after the collision. Since the vertical component before the collision is Vy = V / √2, we can substitute this into the equation:

1/√2 = V'y / (V / √2)

Calculating the Vertical Component After Collision

Rearranging the equation gives us:

V'y = (1/√2) * (V / √2) = V / 2

Horizontal Component Remains Constant

One of the key points to note is that the horizontal component of velocity does not change during the collision, assuming no horizontal forces act on the particle. Therefore:

V'x = Vx = V / √2

Finding the Velocity of Reflection

Now that we have both components of the velocity after the collision, we can find the resultant velocity of the particle after it reflects off the ground:

V' = √(V'x² + V'y²)

Substituting the values we found:

V' = √((V / √2)² + (V / 2)²)

V' = √((V² / 2) + (V² / 4)) = √(3V² / 4) = (V√3) / 2

Determining the Angle of Reflection

The angle of reflection can be calculated using the tangent function, which relates the vertical and horizontal components:

tan(θ') = V'y / V'x

Substituting the values we have:

tan(θ') = (V / 2) / (V / √2) = (1/2) * √2 = √2 / 2

Thus, the angle of reflection θ' can be found as:

θ' = tan⁻¹(√2 / 2)

This corresponds to an angle of 45 degrees, which is consistent with the law of reflection stating that the angle of incidence equals the angle of reflection.

Summary of Results

In summary, after the collision:

  • The velocity of reflection is (V√3) / 2.
  • The angle of reflection is 45 degrees.

This analysis illustrates how the horizontal component remains unchanged while the vertical component is affected by the coefficient of restitution, leading to a clear understanding of the particle's behavior upon striking the ground.

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