Question

1. A block of metal of mass 2kg is resting on
a frictionless floor. It is struck by a jet
releasing water at the rate of 1kg/sec at a
speed of 5ms-1. What will be the initial
acceleration of the block?​

Answer 1

Answer:

The acceleration is  $a = 2.5 \ m/s^2$

Explanation:

From the question we are told that

   The mass of the metal block is $m_b = 2 \ kg$

    The mass flow rate of the water is  $\r m = 1\ kg/s$

    The speed of the water of the water release is $v_w = 5 m/s$

Generally according to the law of conservation of linear momentum

    $p_i = p_f$

Now  $p_i$ is the initial momentum of the system which mathematically represented as

       $p_i = m_w * v_w + m_b * v_b$

Now $m_w$ is the mass of water at the point of contact with the block which can be deduced as $m_w = 1 \ kg$

Now since at initial the block is at rest

       $v_b = 0 \ m/s$

So

      $p_i = 1 * 5$

     $p_i = 5 \ kgm/ s$

And  $p_f$ is the final  momentum of the system which mathematically represented as

     $p_f = m_w * v__{fw} } + m_b * v__{fb}}$

So   $v__{fw} }$ is the final velocity of water which is zero due to the fact that when the water hits the block it losses its momentum and eventually the velocity becomes zero

    So

          $5 = 2 * v__{fb }$

Thus  $v__{fb }} = \frac{5}{2}$

        $v__{fb }} = 2.5 \ m/s$

Thus  

       $p_f = 2.5 * 2$

      $p_f = 5 \ kgm /s$

Now the average momentum change is  

        $p_a = \frac{p_i +p_f}{2}$

       $p_a = \frac{5+5}{2}$

        $p_a =5 kgm/s$

Now the force acting on the block is  

      $F = \frac{p_a }{t}$

and from the question the initial movement of the block took 1s as it is a mass of water moving at a rate of 1kg/s that caused the first movement of the block

  So

        $F= \frac{5}{1}$

       $F= 5 \ N$

Now the acceleration is  

       $a = \frac{F}{m_b}$

=>     $a = \frac{5}{2}$

        $a = 2.5 \ m/s^2$