Question

The 4kg head of a sledge hammer is moving at 6m/s when it strikes a spike dividing it into a log, the duration of impact is 2ms. Find the time average of impact force.

Answer 1

Mass of the hammer, m = 4 kg

Velocity when it strikes, v = 6 m/s

So, momentum before striking = mv = 24 Ns

After striking it comes to rest in 2 ms = 2 × 10-3 s or 0.002 s

Momentum after this time = 0

So, change in momentum, |Δp| = |0 - 24| = 24 Ns

Average force of impact is, F = |Δp|/Δt = 24/0.002

=> F = 12000 N
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Answer 2

Mass of the hammer, m = 4 kg

Velocity when it strikes, v = 6 m/s

So, momentum before striking = mv = 24 Ns

After striking it comes to rest in 2 ms = 2 × 10-3 s or 0.002 s

Momentum after this time = 0

So, change in momentum, |Δp| = |0 - 24| = 24 Ns

Average force of impact is, F = |Δp|/Δt = 24/0.002

=> F = 12000 N

Hope this helps you If so please mark it as Brainliest Answer!! ☺️ ☺️

By soniabikash