Question

Batman is fighting superman. He throws his 5 kg fist at superman with a speed of 9 m/s. Superman stops the punch with a force of 45,000 N. How much time did it take for batman's hand to stop moving?

Answer 1

Answer:

It took $1x10^{-3}$ seconds for batman's hand to stop moving.

Explanation:

Newton' s second law is defined as:

$F = ma$ (1)

Where m is the mass and a is the acceleration.

But it is known that the acceleration is defined as follow by the kinematic equation that corresponds to a Uniformly Accelerated Rectilinear Motion.

$a = \frac{v_{2}-v_{1}}{t}$ (2)

Where $v_{2}$ is the final velocity and $v_{1}$ is the initial velocity

The initial velocity of the fist will be zero ($v_{1} = 0$) since it starts from a state of rest.

Then, equation 2 can be replaced in equation 1

$F = m\frac{v_{2}-v_{1}}{t}$  (3)

Therefore, t can be isolated from equation 3

$t = m\frac{v_{2}-v_{1}}{F}$ (4)

$t = (5Kg)\frac{9m/s-0m/s}{45000Kg.m/s^{2}}$

$t = 1x10^{-3}s$

Hence, it took $1x10^{-3}$ seconds for batman's hand to stop moving.

Answer 2

Answer:

1 × 10^-3 s.

Explanation:

Given:

Mass = 5 kg

V = 9 m/s

Force = 45000 N

Using Newton's law,

Force = mass × acceleration

= (mass × velocity)/time

45000 = (5 × 9)/time

Time = 1 × 10^-3 s.