32. HOTS Question II

Exercise condition:

5 m.

A heavy truck and bike are moving with the same kinetic energy. If the mass of the truck is four times that of the bike, then calculate the ratio of their momenta.

The formula to calculate the Kinetic Energy is,

\(\frac{1}{2}mv^{2}\)
\(mv\)
\(\frac{1}{2}gh^{2}\)
\(mgh\)

Let,

\(m_1,\ m_2\) are the masses of truck and bike and \(v_1,\ v_2\) are the velocities of truck and bike

Ratio of the masses \(=\) \(m_1:4m_2\)

From the given data,

\(Kinetic\ energy\ of\ the\ truck\ =\ Kinetic\ energy\ of\ the\ bike\)

\(\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\)
\(\frac{1}{2}m_{1}gh_{1}=\frac{1}{2}m_{2}gh_{2}\)
\(\frac{1}{2}m_{1}v_{1}=\frac{1}{2}m_{2}v_{2}\)
\(\frac{1}{2}m_{1}gh_{1}^{2}=\frac{1}{2}m_{2}gh_{2}^{2}\)

Substitute the ratio of masses (\(m_1\ =\ 4m_2\)), then

Ratio of velocities (\(v_1\ =\ v_2\)) \(=\) \(:\)

The formula to calculate the Momentum is,

\(mgh\)
\(\frac{1}{2}gh^{2}\)
\(mv\)
\(\frac{1}{2}mv^{2}\)

So the ratio of momenta,

\(\frac{\rho _{truck}}{\rho _{bike}}\) \(=\)

\(\frac{m_{1}gh_{1}}{m_{2}gh_{2}}\)
\(\frac{\frac{1}{2}g_{1}h_{1}^{2}}{\frac{1}{2}g_{2}h_{2}^{2}}\)
\(\frac{\frac{1}{2}m_{1}v_{1}^{2}}{\frac{1}{2}m_{2}v_{2}^{2}}\)
\(\frac{m_{1}v_{1}}{m_{2}v_{2}}\)

Ratio of the momenta (\(\frac{\rho _{truck}}{\rho _{bike}}\)) \(=\) \(:\)

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