Question

A 7500 kg truck traveling at 5m/s east collides with a 1500kg car moving at 20m/s in a direction 30 degrees south of west. After collision, the two vehicles remain tangled together. With what speed and in what direction does the wreckage begin to move?

Answer 1

Answer:

The speed and direction of the wreckage is 2.09 m/s and 52.74°.

Explanation:

Given that,

Mass of truck = 7500 kg

Speed of truck = 5 m/s

Mass of car = 1500 kg

Speed of car = 20 m/s

Angle = 30°

We need to calculate the velocity after collision along east

Using conservation of momentum

$m_{t}v_{t}+m_{c}v_{c}\cos\theta=Mv_{east}$

Put the value into the formula

$7500\times5-1500\times20\cos30=(7500+1500)v_{east}$

$v_{east}=\dfrac{7500\times5-1500\times20\cos30}{7500+1500}$

$v_{east}=1.27\ m/s$

We need to calculate the velocity after collision along north

Using conservation of momentum

$m_{t}v_{t}+m_{c}v_{c}\cos\theta=Mv_{north}$

Put the value into the formula

$7500\times0-1500\times20\sin30=(7500+1500)v_{north}$

$v_{north}=\dfrac{7500\times0-1500\times20\sin30}{7500+1500}$

$v_{north}=-1.67\ m/s$

The resultant velocity is

$v=\sqrt{(v_{east})^2+(v_{north})^2}$

Put the value into the formula

$v=\sqrt{(1.27)^2+(-1.67)^2}$

$v=2.09\ m/s$

We need to calculate the direction

Using formula of direction

$\theta=\tan^{-1}(\dfrac{v_{north}}{v_{east}})$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{|v_{north}|}{|v_{east}|})$

$\theta=\tan^{-1}(\dfrac{1.67}{1.27})$

$\theta=52.74^{\circ}$

Hence, The speed and direction of the wreckage is 2.09 m/s and 52.74°.

Answer 2

The resultant speed (v) in which the wreckage moved is 2.10 m/s and the direction in which the wreckage begins to move is 52.53°

From the given information, we need to find the conservation of momentum in both the east direction and the north direction.

Now, in the east direction, the momentum before the collision is equal to the momentum after the collision.

i.e.

By applying conservation of momentum in the east direction:

$\mathbf{m_ev_e -m_c v_c cos \theta = Mv_{e} }$

$\mathbf{v_{e} =\dfrac{m_ev_e -m_c v_c cos \theta}{M} }$

$\mathbf{v_{e} =\dfrac{(7500\times5) -1500\times20 cos 30^0}{7500+1500} }$

$\mathbf{v_{e} =1.28 \ m/s }$

Also, taking the momentum in the north direction, we have:

$\mathbf{m_nv_n -m_c v_c cos \theta = Mv_{n} }$

$\mathbf{v_{n} =\dfrac{m_nv_n -m_c v_c cos \theta}{M} }$

$\mathbf{v_{n} =\dfrac{(7500\times0) -1500\times20 cos 30^0}{7500+1500} }$

$\mathbf{v_{n} =-1.67 \ m/s }$

Thus, the resultant speed (v) in which the wreckage moved can be expressed by using the Pythagoras rule;

$\mathbf{v ^2 = v_e^2 + v_n^2}$

$\mathbf{v =\sqrt{ v_e^2 + v_n^2}}$

$\mathbf{v =\sqrt{(1.28)^2 + (-1.67)^2}}$

$\mathbf{v = \sqrt{4.4273}}$

v = 2.10 m/s

The direction (i.e. angle θ) where the wreckage begins to move is calculated by the tangent of the angle θ.

i.e.

$\mathbf{tan \theta = \dfrac{v_n}{v_e}}$

$\mathbf{ \theta = tan^{-1}(\dfrac{v_n}{v_e})}$

$\mathbf{ \theta = tan^{-1}(\dfrac{|-1.67|}{|1.28|})}$

$\mathbf{ \theta = tan^{-1}(\dfrac{1.67}{1.28})}$

$\mathbf{ \theta = tan^{-1}(1.3046875)}$

$\mathbf{ \theta = 52.53^0}$

Therefore, we can conclude that the resultant speed (v) in which the wreckage moved is 2.10 m/s and the direction in which the wreckage begins to move is 52.53°

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