JEE Challenger
More from Motion in a Plane
PhyMedAdv 2019 • Paper 2

Calculate Bounce Parameter Alpha for Bouncing Projectile Average Velocity

Consider a ball thrown from the ground at an angle θ\theta to the horizontal with an initial speed u0u_0. In the resulting projectile motion, the magnitude of its average velocity up to the moment it first impacts the ground is V1V_1. Upon hitting the ground, the ball rebounds at the same angle θ\theta, but with a diminished speed of u0/αu_0/\alpha. This bouncing motion continues for a long duration as illustrated in the figure. Given that the magnitude of the average velocity of the ball for the total duration of motion is 0.8V10.8 V_1, find the value of α\alpha.

Question Diagram 1
Official Numerical Answer4

Topics & Concepts

Motion in a PlaneKinematicsProjectile Motion

Step-by-Step Solution

To find the value of α\alpha, we analyze the motion of the ball for both a single bounce and the entire series of bounces.

1. Motion for the First Impact

For the first projectile motion with initial velocity u0u_0 at an angle θ\theta:

  • The time of flight T1T_1 is given by: T1=2u0sin⁡θgT_1 = \frac{2 u_0 \sin\theta}{g}
  • The horizontal range R1R_1 is given by: R1=u02sin⁡2θg=2u02sin⁡θcos⁡θgR_1 = \frac{u_0^2 \sin 2\theta}{g} = \frac{2 u_0^2 \sin\theta \cos\theta}{g}

The magnitude of the average velocity V1V_1 up to the first impact is the net horizontal displacement divided by the time of flight: V1=R1T1=2u02sin⁡θcos⁡θg2u0sin⁡θg=u0cos⁡θV_1 = \frac{R_1}{T_1} = \frac{\frac{2 u_0^2 \sin\theta \cos\theta}{g}}{\frac{2 u_0 \sin\theta}{g}} = u_0 \cos\theta

2. Motion for the Infinite Bounces

After each rebound, the speed decreases by a factor of α\alpha, while the angle θ\theta remains unchanged.

For the nn-th bounce (where n=1,2,3,…n = 1, 2, 3, \dots):

  • The initial speed is un=u0αn−1u_n = \frac{u_0}{\alpha^{n-1}}
  • The time of flight for the nn-th bounce is: Tn=T1αn−1T_n = \frac{T_1}{\alpha^{n-1}}
  • The horizontal range for the nn-th bounce is: Rn=R1α2(n−1)R_n = \frac{R_1}{\alpha^{2(n-1)}}

The total time of motion TtotalT_{\text{total}} for all bounces is the sum of an infinite geometric progression: Ttotal=∑n=1∞Tn=T1(1+1α+1α2+… )=T11−1α=αT1α−1T_{\text{total}} = \sum_{n=1}^{\infty} T_n = T_1 \left(1 + \frac{1}{\alpha} + \frac{1}{\alpha^2} + \dots \right) = \frac{T_1}{1 - \frac{1}{\alpha}} = \frac{\alpha T_1}{\alpha - 1}

Similarly, the total horizontal displacement RtotalR_{\text{total}} for all bounces is: Rtotal=∑n=1∞Rn=R1(1+1α2+1α4+… )=R11−1α2=α2R1α2−1R_{\text{total}} = \sum_{n=1}^{\infty} R_n = R_1 \left(1 + \frac{1}{\alpha^2} + \frac{1}{\alpha^4} + \dots \right) = \frac{R_1}{1 - \frac{1}{\alpha^2}} = \frac{\alpha^2 R_1}{\alpha^2 - 1}

3. Magnitude of Total Average Velocity

The magnitude of the total average velocity VavgV_{\text{avg}} for the entire duration is: Vavg=RtotalTtotal=α2R1α2−1αT1α−1=(R1T1)α2(α−1)α(α2−1)V_{\text{avg}} = \frac{R_{\text{total}}}{T_{\text{total}}} = \frac{\frac{\alpha^2 R_1}{\alpha^2 - 1}}{\frac{\alpha T_1}{\alpha - 1}} = \left(\frac{R_1}{T_1}\right) \frac{\alpha^2 (\alpha - 1)}{\alpha (\alpha^2 - 1)}

Since V1=R1T1V_1 = \frac{R_1}{T_1} and α2−1=(α−1)(α+1)\alpha^2 - 1 = (\alpha - 1)(\alpha + 1), we simplify: Vavg=V1⋅αα+1V_{\text{avg}} = V_1 \cdot \frac{\alpha}{\alpha + 1}

4. Calculation of α\alpha

We are given that Vavg=0.8V1V_{\text{avg}} = 0.8 V_1: αα+1=0.8\frac{\alpha}{\alpha + 1} = 0.8

Solving for α\alpha: α=0.8(α+1)\alpha = 0.8(\alpha + 1) α=0.8α+0.8\alpha = 0.8\alpha + 0.8 0.2α=0.80.2\alpha = 0.8 α=4\alpha = 4

Thus, the value of α\alpha is 4.

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