Asked by Anonymous
A hammer weighing of 1 kg moving with the speed of 10 m/s strikes the head of the nail driving it 10 cm into wall. Neglecting the mass of the nail, calculate:
1) The acceleration during the impact
2) The time interval during the impact
3) The impulse (force X time)
Answers
Answered by
Damon
well, 3 is the easiest of all. Thw impulse is the change of momentum.MV = 1 *10 = 10 kg m/s
average speed during stop = 10/2 = 5 m/s
so
t = time to stop = 0.10 meter/ 5 meters/s = 0.02 second. That is number 2
a = change in v/ time = -10/.02 = -500 m/s^2
that is number 1.
=============================
now you could have one it all this way:
d = Vi t + .5 a t^2
v = Vi + a t
so
0 = 10 + a t
a t = -10
0.10 = 10 t + .5 (-10)t
.1 = 5 t
t = .02 seconds which we knew
then
a t = a(.02)= -10
so a = -500 m/s^2
average speed during stop = 10/2 = 5 m/s
so
t = time to stop = 0.10 meter/ 5 meters/s = 0.02 second. That is number 2
a = change in v/ time = -10/.02 = -500 m/s^2
that is number 1.
=============================
now you could have one it all this way:
d = Vi t + .5 a t^2
v = Vi + a t
so
0 = 10 + a t
a t = -10
0.10 = 10 t + .5 (-10)t
.1 = 5 t
t = .02 seconds which we knew
then
a t = a(.02)= -10
so a = -500 m/s^2
Answered by
shiva
how to find accleration
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