
How much momentum will a dumb-bell of mass \[10\;{\rm{kg}}\] transfer to the floor if it falls from a height of \[80\;{\rm{cm}}\]? Take its downward acceleration to be \[10\;{\rm{m}}{{\rm{s}}^{{\rm{ - 2}}}}\] .
Answer
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Hint:The above problem can be resolved using the mathematical formula of momentum possessed by any object undergoing the linear motion in either direction. The momentum is determined by taking the product of the mass of the given dumb-bell and the velocity with which it was thrown. This magnitude of velocity can be calculated by using the newton's third equation of motion, where the magnitude of initial velocity is zero because the dumb-bell was initially at rest.
Complete step by step answer:
Given:
The mass of dumb-bell is, \[m = 10\;{\rm{kg}}\].
The height of falling is, \[h = 80\;{\rm{cm}} = 80\;{\rm{cm}} \times \dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.8\;{\rm{m}}\].
The downward acceleration is, \[g = 10\;{\rm{m}}{{\rm{s}}^{{\rm{ - 2}}}}\]
The expression for the momentum is given as,
\[P = m \times v\]
Here, v is the velocity of dumb- bell and its value is,
\[{v^2} = {u^2} + 2gh\]
Here, u is the initial velocity and its value is zero as the dumb-bell started from rest.
Substituting the values as,
\[\begin{array}{l}
{v^2} = {u^2} + 2gh\\
{v^2} = {\left( 0 \right)^2} + 2 \times 10\;{\rm{m}}{{\rm{s}}^{{\rm{ - 2}}}} \times 0.8\;{\rm{m}}\\
{{\rm{v}}^2} = 16\\
v = 4\;{\rm{m/s}}
\end{array}\]
The momentum is,
\[\begin{array}{l}
P = m \times v\\
P = 10\,{\rm{kg}} \times 4\;{\rm{m/s}}\\
P = 40\;{\rm{kg - m/s}}
\end{array}\]
Therefore, the momentum of the dumb-bell is \[40\;{\rm{kg - m/s}}\].
Note: Try to remember the formula of the momentum and the concept behind the momentum. As momentum is used to measure the impact of applied force, therefore it has its application to derive Newton's second law of motion. In addition, the basic knowledge of the newton's equation of motion is also required to be taken into consideration, that relates the various variables like speed, acceleration, time involved during the motion of any particular object.
Complete step by step answer:
Given:
The mass of dumb-bell is, \[m = 10\;{\rm{kg}}\].
The height of falling is, \[h = 80\;{\rm{cm}} = 80\;{\rm{cm}} \times \dfrac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.8\;{\rm{m}}\].
The downward acceleration is, \[g = 10\;{\rm{m}}{{\rm{s}}^{{\rm{ - 2}}}}\]
The expression for the momentum is given as,
\[P = m \times v\]
Here, v is the velocity of dumb- bell and its value is,
\[{v^2} = {u^2} + 2gh\]
Here, u is the initial velocity and its value is zero as the dumb-bell started from rest.
Substituting the values as,
\[\begin{array}{l}
{v^2} = {u^2} + 2gh\\
{v^2} = {\left( 0 \right)^2} + 2 \times 10\;{\rm{m}}{{\rm{s}}^{{\rm{ - 2}}}} \times 0.8\;{\rm{m}}\\
{{\rm{v}}^2} = 16\\
v = 4\;{\rm{m/s}}
\end{array}\]
The momentum is,
\[\begin{array}{l}
P = m \times v\\
P = 10\,{\rm{kg}} \times 4\;{\rm{m/s}}\\
P = 40\;{\rm{kg - m/s}}
\end{array}\]
Therefore, the momentum of the dumb-bell is \[40\;{\rm{kg - m/s}}\].
Note: Try to remember the formula of the momentum and the concept behind the momentum. As momentum is used to measure the impact of applied force, therefore it has its application to derive Newton's second law of motion. In addition, the basic knowledge of the newton's equation of motion is also required to be taken into consideration, that relates the various variables like speed, acceleration, time involved during the motion of any particular object.
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