Question

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is

Answer 1

Answer:

a) $v=0.5405\ m.s^{-1}$

b) $v_p'=0.1143\ m.s^{-1}$

Explanation:

Given:

mass of ball, $m_b=4\ kg$

initial speed of the ball, $v_b=10\ m.s^{-1}$

mass of the person, $m_p=70\ kg$

a)

Using the conservation of linear momentum:

When the person catches the ball, assuming that the person catches it with an impact without absorbing the shock.

$m_b.v_b=(m_b+m_p)v$

$4\times 10=(4+70)\times v$

$v=0.5405\ m.s^{-1}$

b)

When the ball hits the person and bounces off with the velocity of $v_b'=8\ m.s^{-1}$.

Using the conservation of linear momentum:

$m_b.v_b+m_p.v_p=m_b.v_b'+m_p.v_p'$

where:

$v_b'=$ final speed of the ball after collision

$v_p'=$ final speed of the person after collision

$v_p=$ initial velocity of the person = 0

putting the respective values in the above eq.

$4\times 10+0=4\times 8+70\times v_p'$

$v_p'=0.1143\ m.s^{-1}$

Answer 2

Answer:

Explanation:

mass of ball, m = 0.4 kg

initial velocity of ball, u = 10 m/s

your mass, M = 70 kg

your initial velocity, U = 0 m/s

(a)

As there is no external force is applied so the linear momentum of the ball and you is conserved.

Let the final velocity of ball and you is V.

So,

Initial Momentum of ball = final momentum of ball and you

m x u = ( M + m) V

0.4 x 10 = ( 70 + 0.4) x V

4 = 70.4 V

V = 0.057 m/s

Thus, your speed is 0.057 m/s after catching the ball.

(b) v' = - 8 m/s

Let V' is the final velocity of you.

Momentum of system before collision = Momentum of system after collision

( 70 + 0.4) x V = m x v' + M x V'

70.4 x 0.045 = - 0.4 x 8 + 70 x V'

4 = - 3.2 + 70 V'

V' = 0.1 m/s

Thus, your speed is 0.1 m/s towards right after hitting by the ball.