Question

A \[1000\,{\text{kg}}\] sports going \[40\,{\text{m/s}}\] slams into an SUV at a stoplight creating an inelastic collision. After the collision, the velocity varies with time according to the following equation: \[v\left( t \right) = 10 - 5{t^{\dfrac{1}{2}}}\] What is the mass of SUV?
A. \[1000\,{\text{kg}}\]
B. \[2500\,{\text{kg}}\]
C. \[3000\,{\text{kg}}\]
D. \[4000\,{\text{kg}}\]
E. \[7000\,{\text{kg}}\]

Answer 1

Hint:Using the equation for time varying velocity of the sports car and SUV system after collision, calculate the velocity just after collision. Use the expression for law of conservation of linear momentum for the system of the sports car and SUV undergoing inelastic collision. Take the velocity of the SUV before collision as zero while substituting the values in the expression for law of conservation of linear momentum.

Formula used:
The expression for law of conservation of linear momentum in an inelastic collision is given by
\[{m_1}{v_1} + {m_2}{v_2} = \left( {{m_1} + {m_2}} \right)v\] …… (1)
Here, \[{m_1}\] and \[{m_2}\] are the masses of the two objects in collision, \[{v_1}\] and \[{v_2}\] are the initial velocities of the two objects in collision and \[v\] is the final velocity of the system of two objects after collision.

Complete step by step answer:
We have given that the mass of the sports car is \[1000\,{\text{kg}}\] and the velocity of the sports car before collision with an SUV is \[40\,{\text{m/s}}\].
\[{m_{Sports}} = 1000\,{\text{kg}}\]
\[{v_{Sports}} = 40\,{\text{m/s}}\]
We have also given that the velocity of the system of sports car and SUV after collision varies with time which is given by
\[v\left( t \right) = 10 - 5{t^{\dfrac{1}{2}}}\] …… (2)
We have asked to calculate the mass of the SUV. Let us first calculate the velocity of the sports car and SUV system just after collision.Substitute \[0\,{\text{s}}\] for \[t\] in equation (2).
\[v\left( {0\,{\text{s}}} \right) = 10 - 5{\left( {0\,{\text{s}}} \right)^{\dfrac{1}{2}}}\]
\[ \Rightarrow v\left( {0\,{\text{s}}} \right) = 10\,{\text{m/s}}\]
Hence, the velocity of the sports car and SUV system just after the collision is \[10\,{\text{m/s}}\].

Since the SUV is at the stoplight before collision in rest position. The initial velocity of SUV before collision is zero.
\[{v_{SUV}} = 0\,{\text{m/s}}\]
We can calculate the mass of the SUV using law of conservation of linear momentum for inelastic collision.
Rewrite equation (1) for the inelastic collision between the sports car and SUV.
\[{m_{Sports}}{v_{Sports}} + {m_{SUV}}{v_{SUV}} = \left( {{m_{Sports}} + {m_{SUV}}} \right)v\]
Substitute \[1000\,{\text{kg}}\] for \[{m_{Sports}}\], \[40\,{\text{m/s}}\] for \[{v_{Sports}}\], \[0\,{\text{m/s}}\] for \[{v_{SUV}}\] and \[10\,{\text{m/s}}\] for \[v\] in the above equation.
\[\left( {1000\,{\text{kg}}} \right)\left( {40\,{\text{m/s}}} \right) + {m_{SUV}}\left( {0\,{\text{m/s}}} \right) = \left( {1000\,{\text{kg}} + {m_{SUV}}} \right)\left( {10\,{\text{m/s}}} \right)\]
\[ \Rightarrow 40000 = \left( {1000\,{\text{kg}} + {m_{SUV}}} \right)\left( {10\,{\text{m/s}}} \right)\]
\[ \Rightarrow 4000 = 1000 + {m_{SUV}}\]
\[ \Rightarrow {m_{SUV}} = 4000 - 1000\]
\[ \therefore {m_{SUV}} = 3000\,{\text{kg}}\]
Therefore, the mass of the SUV is \[3000\,{\text{kg}}\].

Hence, the correct option is C.

Note:The students may think that the velocity of the sports car and SUV system at zero seconds is not the velocity after collision because at the zero second both the cars must be in the collision process. But the students should keep in mind that in question we have given that they have given expression for velocity for the time varying velocity after collision. Hence, we have calculated the velocity of the system at zero second after collision.