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A 30.0 kg hammer, moving with speed a 20...

A `30.0 kg` hammer, moving with speed a `20.0ms^(-1)` strikes a steel spike `2.30 cm` in diameter. The hammer rebounds with speed `10.0ms^(-1)` after `0.110s.` What is the average strain in the spike during the impact.?

A

`1.97xx10^(7) N//m^(2)`

B

`3.2xx10^(7) N//m^(2)`

C

`4.6xx10^(7) N//m^(2)`

D

`8.2xx10^(7) N//m^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
The force acting on the hammer changes its momentum according to `mv_(1) + vec(F) (Delta t) = mv_(f)` so
`|vec(F)| = (m|v_(f) - v_(i)|)/(Delta t)`, Hence, `|vec(F)| = 8.18xx10^(3)N`
By Newton's third law, that is also the magnitude of the average force exerted on the spike by the hammer during the blow thus, the stress in the spike is
Stress =
`(F)/(A) = (8.18xx10^(3) N)/(pi (0.023m^(2))//4) = 1.97xx10^(7)N//m^(2)`
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