1. A 601 kg car stopped at an intersection is rear-ended by a 1880 kg truck moving with a speed of 14.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.
_____________m/s (car)
_____________m/s (truck)
2.An explosion breaks an object into two pieces, one of which has 1.80 times the mass of the other. If 7800 J were released in the explosion, how much kinetic energy did each piece acquire?
heavier piece
_________J
lighter piece
____________ J
1. M1 = 601 kg, V1 = 0.
M2 = 1880 kg, V2 = 14.5 m/s.
M1*V1+M2*V2 = M1*V3+M2*V4.
601*0+1880*14.5 = 601*V3+1880*V4,
Eq1: 601V3+1880V4 = 27,260,
V3 = ((M1-M2)V1 + 2M2*V2)/(M1+M2).
V3 = ((601-1880)0 + 3760*14.5)/(601+1880) = 54,520/2481 = 22 m/s.
In Eq1, replace V3 with 22 and solve for V4.
To solve these problems, we can use the principles of conservation of momentum and conservation of kinetic energy.
1. Final Speeds of Car and Truck after Collision:
To find the final speeds of the car and truck after the collision, we need to use the conservation of momentum equation.
According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
Total momentum before collision = Total momentum after collision
Momentum (P) = mass (m) x velocity (v)
For the car: m1 = 601 kg (mass), v1 = 0 (initial velocity), v1' (final velocity - to be found)
For the truck: m2 = 1880 kg (mass), v2 = 14.5 m/s (initial velocity), v2' (final velocity - to be found)
Using the conservation of momentum equation:
m1v1 + m2v2 = m1v1' + m2v2'
601 kg x 0 + 1880 kg x 14.5 m/s = 601 kg x v1' + 1880 kg x v2'
Simplifying the equation:
1880 kg x 14.5 m/s = 601 kg x v1' + 1880 kg x v2'
Now, solve for the final velocities:
v1' + v2' = (1880 kg x 14.5 m/s) / 601 kg
v1' + v2' = 45.338 m/s
We don't have enough information in the question to determine the final velocities of each vehicle individually. However, we know their sum is 45.338 m/s.
2. Kinetic Energy of Each Piece after the Explosion:
To find the kinetic energy acquired by each piece after the explosion, we can use the conservation of kinetic energy.
According to the law of conservation of kinetic energy, the total kinetic energy before the explosion is equal to the total kinetic energy after the explosion.
Total kinetic energy before explosion = Total kinetic energy after explosion
The formula for kinetic energy is K.E. = 0.5 x mass x velocity^2.
Let's assume the mass of the smaller object is 'm' and the mass of the larger object is 1.8m.
Based on this, the formula for kinetic energy becomes:
K.E. (smaller piece) = 0.5 x m x v^2
K.E. (larger piece) = 0.5 x (1.8m) x v^2
Given that a total of 7800 J of kinetic energy was released in the explosion:
K.E. (smaller piece) + K.E. (larger piece) = 7800 J
Substituting the kinetic energy formulas:
0.5 x m x v^2 + 0.5 x (1.8m) x v^2 = 7800 J
Simplifying the equation:
0.5m v^2 + 0.9m v^2 = 7800 J
1.4m v^2 = 7800 J
Now, solve for the kinetic energy of each piece:
0.5m v^2 = (7800 J) / 1.4
v^2 = (7800 J) / (1.4m)
The kinetic energy of the heavier piece:
K.E. (heavier piece) = 0.5 x (1.8m) x v^2
K.E. (heavier piece) = 0.9m x v^2
Substituting the value of v^2:
K.E. (heavier piece) = 0.9m x [(7800 J) / (1.4m)]
Simplifying the equation:
K.E. (heavier piece) = 0.9 x (7800 J) / 1.4
Similarly, the kinetic energy of the lighter piece:
K.E. (lighter piece) = 0.5 x m x v^2
K.E. (lighter piece) = 0.5m x v^2
Substituting the value of v^2:
K.E. (lighter piece) = 0.5m x [(7800 J) / (1.4m)]
Simplifying the equation:
K.E. (lighter piece) = 0.5 x (7800 J) / 1.4
After simplification, the expressions for the kinetic energy of each piece will depend on the mass 'm' of the smaller piece. We would need additional information to determine the specific values.