Question

Find the speed at which Superman (mass=78.0 kg) must fly into a train (mass = 17863 kg) traveling at 75.0 km/hr to stop it. Running into the train at that speed would severely damage both train and passengers. Calculate the minimum time Superman must take to stop the train, if the passengers experience an average horizontal force of 0.580 their own weight. How far does the train then travel while being slowed to a stop?

Answer 1

Answer:

Explanation:

mass of superman, m = 78 kg

mass of train, M = 17863 kg

(a)

initial velocity of train, U = 75 km/h = 20.83 m/s

let u is the velocity of superman.

By using the conservation of momentum

m x u = M x V

78 x u = 17863 x 75

u = 17175.96 km/h = 4771.1 m/s

(b)

a = - 0.58 x 9.8 = - 5.684 m/s²

by first equation of motion

v = U + at

0 = 20.83 - 5.684 t

t = 3.66 s

use second equation of motion

s = Ut + 0.5 x at²

s = 20.83 x 3.66 - 0.5 x 5.684 x 3.66 x 3.66

s = 76.24 - 38.1

s = 38.1 m

Answer 2

Answer with Explanation:

We are given that

Mass of superman=m=78 kg

Mass of train=m'=17863 kg

Speed of train=u'=75 km/h=$75\times \frac{5}{18}=20.8 m/s$

$1 km/h=\frac{5}{18} m/s$

Let initial speed of superman=u

Momentum=mv

Using the formula

$78u=17863\times 75$

$u=\frac{17863\times 75}{78}$

$u=17175.9 km/h$

Average horizontal force=0.58

Deceleration $a=-0.58\times 9.8=-5.68 m/s^2$

Final speed of train=v'=0

$v=u+at$

Using the formula

$0=20.8-5.68t$

$5.68t=20.8$

$t=\frac{v'-u'}{a}=\frac{0-20.8}{-5.68}$

$t=3.66 s$

$v^2-u^2=2as$

Using the formula

$0-(20.8)^2=2(-5.68)a$

$(20.8)^2=2(5.68)s$

$s={\frac{(20.8)^2}{2(5.68)}=38.1 m$