Answer to Question #107219 in Mechanics | Relativity for mira

Question

A 1 215.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 8 900.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.

What is the velocity of the truck right after the collision?



What is the change in mechanical energy of the car–truck system in the collision?

Answer 1

Since all movements were made along the same direction, we can use momentum modules with plus sign in east direction. Before the collision, the momentum of the car was

(1) $P_{c0}=m_c \cdot V_{c0}$ where $m_c=1215.0 kg$ , $V_{c0}=25.000 ms^{-1}$

The momentum of the truck was

(2) $P_{t0}=m_t \cdot V_{t0}$ where $m_t=8900.0 kg$ , $V_{t0}=20.000 ms^{-1}$ .

After the collision, the momentum of the car is

(3) $P_{c1}=m_c\cdot V_{c1}$ where $V_{c1}=18.000 ms^{-1}$

The law of conservation of momentum states

(4) $P_{c0}+P_{t0}=P_{c1}+P_{t1}$ where $P_{t1}$ - the momentum of truck after crashes.

Thus we have

(5) $P_{t1}=P_{t0}+P_{c0}-P_{c1}=m_t \cdot V_{t0}+ m_c \cdot (V_{c0} -V_{c1})$. To find truck velocity after crashes we taken into account $P_{t1}=m_t\cdot V_{t1}$ and dived (5) by $m_t$ , we have

(6) $V_{t1}=V_{t0}+\frac{m_c}{m_t}\cdot (V_{c0}-V_{c1})=[20 +\frac{1215}{8900}\cdot (25-18)]ms^{-1}= 20.956ms^{-1}$

The initial kinetic energy is

(7) $E_i=\frac{m_c\cdot V_{c0}^2}{2}+\frac{m_t\cdot V_{t0}^2}{2}$

The final kinetic energy is

(8) $E_f=\frac{m_c\cdot V_{c1}^2}{2}+\frac{m_t\cdot V_{t1}^2}{2}$

According to (6) and (2), the change in the speed of the truck is quite small, so care must be taken when calculating the energy difference. With such small changes in calculating the difference, a large loss of accuracy is possible. Find the energy difference as possibly analytically and separately for car and truck.

(9) $\Delta E_c=E_{cf}-E_{ci}=\frac{m_c\cdot V_{c1}^2}{2}-\frac{m_c\cdot V_{c0}^2}{2}=\frac{m_c}{2}(V_{c1}^2-V_{c0}^2)=\\=\frac {m_c}{2}(V_{c1}+V_{c0})\cdot(V_{c1}-V_{c0})=\frac{1215\cdot 43\cdot (-7)}{2}=-182857.50 J$

(10) $\Delta E_t=E_{tf}-E_{ti}=\frac{m_t\cdot V_{t1}^2}{2}-\frac{m_t\cdot V_{t0}^2}{2}=\frac{m_t}{2}(V_{t1}^2-V_{t0}^2)=\\=\frac {m_t}{2}(V_{t1}+V_{t0})\cdot(V_{t1}-V_{t0})$ .

Substitute in (10) formula (6) for $V_{t1}$

(11) $\Delta E_t=\frac {m_t}{2}(2V_{t0}+\frac{m_c}{m_t}\cdot (V_{c0}-V_{c1}))\cdot \frac{m_c}{m_t}(V_{c0}-V_{c1}))=\\=m_c(V_{t0}+\frac{m_c}{2m_t}\cdot (V_{c0}-V_{c1}))\cdot (V_{c0}-V_{c1})=1215(20+\frac{1215}{2\cdot 8900}7)\cdot 7=174163.76 J$

Thus $\Delta E_k=\Delta E_t+\Delta E_c=174163.76 J -182857.50 J=-8693.7 J$

Answers: The velocity of the truck right after the collision is $20.956ms^{-1}$ . The change in mechanical energy of the car–truck system in the collision is $-8693.7 J$