Question

A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A) What is the player's speed afterward if the ball is thrown at 15 m/s relative to the ground?B) What is the player's speed afterward if the ball is thrown at 15 m/s relative to the player?

Answer 1

Answer:

4.53482 m/s

4.506 m/s

Explanation:

$m_1$ = Mass of player = 75 kg

$v_1$ = Initial velocity of player = 4.6 m/s

$m_2$ = Mass of ball = 0.47 kg

$v_1$ = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

$(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s$

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

$(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s$

The player's speed is 4.506 m/s

Answer 2

The player's speed will be:

(A) 4.535 m/s

(B) 4.506 m/s

Given values,

Mass,

• $m_1 = 75 \ kg$
• $m_2 = 0.47 \ kg$

Velocity,

• $v_1 = 4.6 \ m/s$
• $v_2 = 15 \ m/s$

(A)

We know,

→ $(m_1+m_2)v_1 = m_1 v+m_2 v_2$

or,

→                  $v_1 = \frac{(m_1+m_2)v_1-m_2 v_2}{m_1}$

By putting the values, we get

                        $= \frac{(75+0.47)4.6-0.47\times 15}{75}$

                        $= 4.535 \ m/s$

(B)

By using the equation of momentum, we get

→ $(m_1+m_2)v_1= m_1 v+m_2(v_2+v_1)$

or,

→                   $v = \frac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}$

By putting the values, we get

                        $= \frac{(75+0.47)4.6-0.47(15+4.6)}{75}$

                        $= 4.506 \ m/s$

Thus the above response is correct.  

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