You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is
negligible friction between your feet and the ice. A friend throws you a 0.400-kg ball that is
traveling horizontally at 10.0 m/s. Your mass is 70.0 kg. (a) If you catch the ball, with what
speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so
afterward it is moving horizontally at 8.0 m/s in the opposite direction, what is your speed after
the collision?
Identify: Apply conservation of momentum to the system of you and the ball. In part (a) both
objects have the same final velocity.
Set Up: Let +x be in the direction the ball is traveling initially.
0.400 kg mA

(ball).
70.0 kg mB

(you).
Execute:
(a)
P P 1 2 x x

gives
2
(0.400 kg)(10.0 m/s) (0.400 kg 70.0 kg)   v
and
2
v  0.0568 m/s
.
(b)
P P 1 2 x x

gives
2
(0.400 kg)(10.0 m/s) (0.400 kg)( 8.00 m/s) (70.0 kg) B    v
and
2 0.103 m/s B
v 
.
Evaluate: When the ball bounces off it has a greater change in momentum and you acquire a
greater final speed.

Respuesta :

Answer:

a) [tex]v=0.5405\ m.s^{-1}[/tex]

b) [tex]v_p'=0.1143\ m.s^{-1}[/tex]

Explanation:

Given:

mass of ball, [tex]m_b=4\ kg[/tex]

initial speed of the ball, [tex]v_b=10\ m.s^{-1}[/tex]

mass of the person, [tex]m_p=70\ kg[/tex]

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.

[tex]m_b.v_b=(m_b+m_p)v[/tex]

[tex]4\times 10=(4+70)\times v[/tex]

[tex]v=0.5405\ m.s^{-1}[/tex]

b)

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

Using the conservation of linear momentum:

[tex]m_b.v_b+m_p.v_p=m_b.v_b'+m_p.v_p'[/tex]

where:

[tex]v_b'=[/tex] final speed of the ball after collision

[tex]v_p'=[/tex] final speed of the person after collision

[tex]v_p=[/tex] initial velocity of the person = 0

putting the respective values in the above eq.

[tex]4\times 10+0=4\times 8+70\times v_p'[/tex]

[tex]v_p'=0.1143\ m.s^{-1}[/tex]

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.