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

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

Step-by-step 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 )} = (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 = (p_i +p_f)/(2)`

`p_a = (5+5)/(2)`

`p_a =5 kgm/s`

Now the force acting on the block is

`F = (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= (5)/(1)`

`F= 5 \ N`

Now the acceleration is

`a = (F)/(m_b)`

=>
`a = (5)/(2)`

`a = 2.5 \ m/s^2`