Question

A ball is dropped on the floor from a height of 10 m it rebounds to a height of 2.5m. If the ball is in contact with the floor for 0.01 seconds, the average acceleration during contact will be: 

1. 700 m/s²
2. 1400 m/s2
3. 2100 m/s2
4. 2800 m/s2

Answer 1

Correct Answer - Option 3 : 2100 m/s2

CONCEPT:

Acceleration:

• When the velocity of a moving object changes either in magnitude or in direction or in both, then the body is said to be in acceleration.
• The rate of change of velocity is called acceleration.

\(⇒ a =\frac{Δ v}{Δ t}\)

Where Δv = change in velocity and Δt = time

Equation of Kinematics:

• These are the various relations between u, v, a, t, and s for the particle moving with uniform acceleration where the notations are used as:
• Equations of motion can be written as

⇒ V = U + at

\(⇒ s =Ut+\frac{1}{2}{at^{2}}\)

⇒ V2 = U2+ 2as

where, U = Initial velocity, V = Final velocity, g = Acceleration due to gravity, t = time, and h = height/Distance covered

CALCULATION:

The third equation of motion is given as,

⇒ V2 = U2 + 2as     -----(1)

In case 1, (U₁ = 0 m/s, s₁ = 10 m, a₁ = g = 9.8 m/s²)

\(⇒ V_{1}^{2}=U_{1}^{2}+2gs_{1}\)

\(⇒ V_{1}^{2}=0^{2}+(2\times9.8\times10)\)

⇒ V₁ = 14 m/s (Downward)

In case 2 when the ball rebounds, (V2 = 0 m/s, s2 = 2.5 m, a2 = -g = -9.8 m/s2)

\(⇒ V_{2}^{2}=U_{2}^{2}-2gs_{2}\)

\(⇒ 0^{2}=U_{2}^{2}-(2\times9.8\times2.5)\)

⇒ U2 = 7 m/s (Upward)

• So the change in velocity during contact is given as,

⇒ ΔV = U₂ - V₁

⇒ ΔV = 7 - (-14)

⇒ ΔV = 7 + 14

⇒ ΔV = 21 m/s

And,

⇒ Δt = 0.01 sec

So the acceleration during contact,

\(⇒ a =\frac{ΔV}{Δ t}\)

\(⇒ a =\frac{21}{0.01}\)

⇒ a = 2100 m/s²

• Hence, option 3 is correct.