Question

Steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m.

Answer 1

Answer:

Vb= 5.42 m/s

Vc= 5.33 m/s

$a=-1125\ m/s^2$

Δ L= 0.43 mm

Explanation:

a)  A to B

Initial velocity Va = 0 m/s

As we know that

$V^2=U^2+2aS$

$V_b^2=2\times 9.81\times 1.5$

Vb= 5.42 m/s

b)

C to D

Vd= 0 m/s

$0^2=V_c^2-2\times 9.81\times 1.45$

Vc= 5.33 m/s

c)

Here

Vb= 5.42 m/s

Vc= 5.33 m/s

We know that

Acceleration is the rate of change of velocity

$a=\dfrac{V_c-V_b}{t}$

$a=\dfrac{5.33-5.42}{8\times 10^{-5}}$

$a=-1125\ m/s^2$

d)

Now length compressed given as

$V_c^2=V_b^2+2a\Delta L$

$5.33^2=5.42^2-2\times 1125\times \Delta L$

  Δ L= 0.43 mm