Question

A hammer weighing \[2.5\,{\text{kg}}\] moving with a speed of \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] strikes the head of the nail driving it \[10\,{\text{cm}}\] into the nail wall. The acceleration during impact and the impulse imparted to wall will be respectively
A. \[ - 5\,{\text{m}} \cdot {{\text{s}}^{ - 2}},\,2.5\,{\text{N}} \cdot {\text{s}}\]
B. \[ - 10\,{\text{m}} \cdot {{\text{s}}^{ - 2}},\,5\,{\text{N}} \cdot {\text{s}}\]
C. \[ - 7.5\,{\text{m}} \cdot {{\text{s}}^{ - 2}},\,3.5\,{\text{N}} \cdot {\text{s}}\]
D. \[ - 6\,{\text{m}} \cdot {{\text{s}}^{ - 2}},\,4\,{\text{N}} \cdot {\text{s}}\]

Answer 1

Hint:Use the expression for the kinematic equation in terms of displacement of the object. Using this formula, calculate the acceleration of the hammer during its impact. Also use the formula for impulse imparted to an object in terms of change in momentum of the object. Using this formula, calculate the impulse imparted by the hammer to the wall.

Formulae used:
The kinematic equation for the final velocity \[v\] in terms of displacement is
\[{v^2} = {u^2} + 2as\] …… (1)
Here, \[u\] is initial velocity of the object, \[a\] is acceleration of the object and \[s\] is displacement of the object.
The impulse \[J\] imparted to an object is given by
\[J = m\left( {v - u} \right)\] …… (2)
Here, \[m\] is mass of the object, \[v\] is final velocity of the object and \[u\] is initial velocity of the object.

Complete step by step answer:
We have given that the mass of the hammer is \[2.5\,{\text{kg}}\].
\[m = 2.5\,{\text{kg}}\]
The initial speed of the hammer is \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\].
\[u = 1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
As the hammer stops after it strikes the head of the nail. Hence, the final speed of the hammer is zero.
\[v = 0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\]
The displacement of the nail into the wall is \[10\,{\text{cm}}\].
\[s = 10\,{\text{cm}}\]
We have asked to calculate the acceleration of the hammer during the impact.Rearrange equation (1) for acceleration of the hammer during the impact.
\[a = \dfrac{{{v^2} - {u^2}}}{{2s}}\]

Substitute \[0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[v\], \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[u\] and \[10\,{\text{cm}}\] for \[s\] in equation (1).
\[a = \dfrac{{{{\left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2} - {{\left( {1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2}}}{{2\left( {10\,{\text{cm}}} \right)}}\]
\[ \Rightarrow a = \dfrac{{{{\left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2} - {{\left( {1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)}^2}}}{{2\left[ {\left( {10\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]}}\]
\[ \Rightarrow a = - 5\,{\text{m}} \cdot {{\text{s}}^{ - 2}}\]
Therefore, the acceleration of the hammer during the impact is \[ - 5\,{\text{m}} \cdot {{\text{s}}^{ - 2}}\].

Let us now calculate the impulse imparted by hammer to the wall.Substitute \[2.5\,{\text{kg}}\] for \[m\], \[0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[v\] and \[1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}\] for \[u\] in equation (2).
\[J = \left( {2.5\,{\text{kg}}} \right)\left[ {\left( {0\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right) - \left( {1\,{\text{m}} \cdot {{\text{s}}^{ - 1}}} \right)} \right]\]
\[ \Rightarrow J = - 2.5\,{\text{kg}} \cdot {\text{m}} \cdot {{\text{s}}^{ - 1}}\]
\[ \therefore J = - 2.5\,{\text{N}} \cdot {\text{s}}\]
Hence, the impulse imparted to the wall by the hammer is \[2.5\,{\text{N}} \cdot {\text{s}}\].The negative sign indicates that the momentum of the hammer decreases.

Hence, the correct option is A.

Note: One can also solve the same question by another method. One can calculate the acceleration of the hammer during the impact and time for which the hammer is in contact with the head of the nail using kinematic equations. Then one can calculate the impulse imparted to the wall by the hammer in terms of force on the nail and the time for which hammer is in contact with the nail head.