With what impulse does a 0.50-kg newspaper have to be thrown to give it a velocity of \({\bf{3}}{\bf{.0}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.\\} {\bf{s}}}\)?

Impulse can be defined as the shift in the object's momentum. For a higher value of the applied force, the value of impulse is also higher.

The mass of the newspaper is \(m = 0.50\;{\rm{kg}}\).

The final velocity of the newspaper is \({v_f} = 3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

The initial velocity of the newspaper is \({v_i} = 0\).

The impulse exerted on the newspaper can be calculated as shown below:

\(\begin{array}{l}I = m\left( {{v_f} - {v_i}} \right)\\I = \left( {0.50\;{\rm{kg}}} \right)\left[ {\left( {3.0\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right) - \left( 0 \right)} \right]\\I = 1.50\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

Thus, the impulse exerted on the newspaper is \(1.50\;{{{\rm{kg}} \cdot {\rm{m}}} \mathord{\left/{\vphantom {{{\rm{kg}} \cdot {\rm{m}}} {\rm{s}}}} \right.\\} {\rm{s}}}\).