A truck of 5000kg hits a car of 1500kg at an intersection. After the collision, both vehicles remained glued together and slid over a distance of 30 meters before coming to a halt. Knowing that the coefficient of kinetic friction between both vehicles and the street is 0.700, find the velocity of the truck before the collision.
The velocity immediately after contact is found by
1500*0+5000V=6500V'
So the energy after the collision in motion is 1/2 (6500)V'^2
that must equal (6500)g*mu*distance
1/2 6500 V'^2=6500 g*.7*30
or
V'^2=2g*.7*30
But V'=5000/6500 V
now put that in, and solve for V
how do you find this?
1500*0+5000V=6500V'
Thank you for your time
To find the velocity of the truck before the collision, we can use the principle of conservation of momentum.
The momentum before the collision is equal to the momentum after the collision. Momentum is calculated as the product of mass and velocity.
Let's assume the initial velocity of the truck is \( V_t \) (in m/s) and the initial velocity of the car is \( V_c \). Since the two vehicles remained glued together after the collision, their final velocity can be considered the same.
The momentum before the collision is given by:
\( Momentum_{before} = (mass_{truck} \times V_{truck}) + (mass_{car} \times V_{car}) \)
\( Momentum_{before} = (5000 \, kg \times V_t) + (1500 \, kg \times V_c) \)
The momentum after the collision is given by:
\( Momentum_{after} = (mass_{combined} \times V_{final}) \)
\( Momentum_{after} = ((5000 \, kg + 1500 \, kg) \times V_{final}) \)
Since the momentum before the collision is equal to the momentum after the collision, we can equate the two:
\( (5000 \, kg \times V_t) + (1500 \, kg \times V_c) = ((5000 \, kg + 1500 \, kg) \times V_{final}) \)
Simplifying this equation, we get:
\( 5000 \, kg \times V_t + 1500 \, kg \times V_c = 6500 \, kg \times V_{final} \)
We also know that the combined vehicles slid over a distance of 30 meters before coming to a halt. Using the equation for work done against friction, we can find the force of friction:
\( Work = Force \times Distance \)
The work done against friction is given by:
\( Work = Force_{friction} \times Distance \)
\( Work = \text{friction coefficient} \times \text{Normal force} \times Distance \)
The normal force is equal to the weight, which is calculated as the mass multiplied by the acceleration due to gravity (\( 9.8 \, m/s^2 \)).
So, the force of friction is given by:
\( Force_{friction} = \text{friction coefficient} \times (\text{mass}_{truck} + \text{mass}_{car}) \times \text{acceleration due to gravity} \)
\( Force_{friction} = 0.7 \times (5000 \, kg + 1500 \, kg) \times 9.8 \, m/s^2 \)
Now, using Newton's second law (\( F = ma \)), we can find the acceleration of the combined vehicles:
\( Force_{friction} = \text{mass}_{combined} \times \text{acceleration} \)
\( 0.7 \times (5000 \, kg + 1500 \, kg) \times 9.8 \, m/s^2 = (5000 \, kg + 1500 \, kg) \times \text{acceleration} \)
Now, we can use the kinematic equation for linear motion to find the final velocity of the combined vehicles:
\( V_{final}^2 = V_{initial}^2 + 2 \times \text{acceleration} \times \text{distance} \)
\( V_{final}^2 = V_{initial}^2 + 2 \times \frac{{0.7 \times (5000 \, kg + 1500 \, kg) \times 9.8 \, m/s^2 \times 30 \, m}}{{(5000 \, kg + 1500 \, kg)}} \)
Rearranging the equation, we get:
\( V_{initial} = \sqrt{{V_{final}^2 - 2 \times \frac{{0.7 \times (5000 \, kg + 1500 \, kg) \times 9.8 \, m/s^2 \times 30 \, m}}{{(5000 \, kg + 1500 \, kg)}}}} \)
Now, substitute the known values into the equation and solve for \( V_{initial} \) to find the velocity of the truck before the collision.