Question

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

Answer 1

It Seems like you are asking Impact Force over the given average Time interval.

Final Answer : $\red {12000N}$ opposite to motion of Hammer.

Steps:
1) We are assuming Impact Force on her is constant over the given time interval.
Given:
Mass of Hammer ,m: 4 kg
Initial velocity of hammer, u = 6m/s
Final velocity is assumed to reach zero (when it is cut) , v = 0
Time Interval, $\Delta t = 2 *10^{-3}s$

2) By Impulse Momentum Theorem,
$F \Delta t = \Delta p \\ => F *2*10^{-3} = 4*(0-6) \\ => F = -12 * 10^{3} N = -12 KN$

Hence, Impact Force will act opposite to motion of Hammer by the spike of Magnitude 12KN.

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