Question

A sharp object of mass 0.5 kg travelling horizontally at a velocity of 100 m/s strikes a stationary wooden block of mass 2.5kg.Therefore the two move together in the same straight line. Calculate the total momentum before and after the impact. Also calculate the velocity of the combined system

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Answer 1

Solution :-

• Mass of the object = 0.5 kg
• Initial velocity of the object = 100 m /s
• Mass of the wooden block = 2.5 kg
• Initial velocity of the block = 0 m/s

(As the block was stationary )

Initial momentum of the system  

$\implies\mathtt{ P_i= P_o + P_b }$

here ,

• Pi = Initial momentum of the system  
• Po = momentum of object
• P b = momentum of block

$\implies\mathtt{ Pi = m_o \times v_o + m_b \times v_b}$

here ,

• mₒ = mass of object
• vₒ = velocity of object
• m b = mass of block
• v b = velocity of block

Now let's substitute the given values in the above equation ,

$\implies\mathtt{ P_i= 0.5\times 100+ 2.5 \times 0}$

$\implies\mathtt{ P_i= 50 \: kg \dfrac{m}{s} }$

The momentum of the system before collision is 50 kg m /s

After the sharp object collides with the stationary wooden block both move with a common velocity in the same direction as that of the object

As no external forces acting on the system total momentum of the system is conserved.

Hence, we can apply the law of conservation of momentum in order to solve this question

This means that the initial momentum of the system will be equal to the final momentum

As per the law of conservation of momentum,

$\implies\mathtt{P_i = P _ f = 50 \dfrac{kgm}{s} }$

The final momentum of the system is 50 kg m /s

Now that we have the value of final momentum we can easily find the common velocity of the system after the collision.

$\implies\mathtt{ P_f = (m_o+ m_b )v}$

$\implies\mathtt{ 50 = (m_o+ m_b )v}$

$\implies\mathtt{ 50 = (0.5 + 2.5 )v}$

$\implies\mathtt{ 50 = 3v}$

$\implies\mathtt{ v = \dfrac{50}{3} }$

$\implies\mathtt{ \red{v = 16.7 \dfrac{m}{s} }}$

The common velocity of the entire system is 16.7 m /s.