A toy car with a mass of 120 g moves to the right with a speed of 0.56 m/s. A small child drops a 25.0-g piece of clay onto the car. The clay sticks to the car and the car continues to the right. What is the change in speed of the car? Consider the frictional force between the car and the ground to be negligible. Answer in m/s.
To solve this problem I tried to use m1v1=m2v2.
I plugged in 120(0.56) = (120+25)v2 = 120(0.56)=145v2 = 0.46 = v2
Then I tried to solve for the change in speed by taking v1-v2 = 0.56-0.46 = 0.1m/s.
This answer is wrong. Please assist.
To correctly solve this problem, you need to use the principle of conservation of momentum. According to this principle, the total momentum before the clay is dropped is equal to the total momentum after the clay sticks to the car.
The momentum of an object is given by the product of its mass and velocity. Let's denote the toy car as object 1 and the clay as object 2.
Initial momentum of the toy car (m1v1) = (mass of the car) * (velocity of the car)
Momentum of the clay (m2v2) = (mass of the clay) * (velocity of the clay)
Total initial momentum = Initial momentum of the toy car + Momentum of the clay
After the clay sticks to the car, the final momentum is the momentum of the car with the clay attached.
Final momentum of the car with clay (m1+m2) * v'
Using the principle of conservation of momentum, we can equate the initial and final momenta:
(mass of the car) * (velocity of the car) + (mass of the clay) * (velocity of the clay) = (mass of the car + mass of the clay) * (final velocity of the car with clay)
Plugging in the given values:
(0.120 kg) * (0.56 m/s) + (0.025 kg) * (0 m/s) = (0.120 kg + 0.025 kg) * (final velocity of the car with clay)
0.0672 kg*m/s = 0.145 kg * (final velocity of the car with clay)
Now, solve for the final velocity of the car with clay:
(final velocity of the car with clay) = 0.0672 kg*m/s / 0.145 kg
(final velocity of the car with clay) ≈ 0.463 m/s
The change in speed of the car is the final velocity of the car with clay minus the initial velocity of the car:
Change in speed = 0.463 m/s - 0.56 m/s
Change in speed ≈ -0.097 m/s
Thus, the change in speed of the car is approximately -0.097 m/s.
To solve this problem, we can use the principle of conservation of momentum. The principle states that the total momentum of a system is conserved if no external forces are acting on it.
In this case, before the clay is dropped, the momentum of the system is the product of the mass of the car and its velocity:
m1 * v1 = (120 g) * (0.56 m/s)
Now, when the clay is dropped onto the car and sticks to it, the total mass of the system becomes the sum of the masses of the car and the clay:
m1 + m2 = (120 g) + (25 g)
The velocity of the combined system after the clay is dropped can be represented as v2.
So, we have:
(m1 + m2) * v2 = (m1 * v1) + (m2 * v2)
Substituting the given values:
(120 g + 25 g) * v2 = (120 g * 0.56 m/s) + (25 g * v2)
Simplifying:
(145 g) * v2 = (67.2 g m/s) + (25 g * v2)
Now, let's convert the masses to kilograms (1 kg = 1000 g):
(0.145 kg) * v2 = (0.0672 kg m/s) + (0.025 kg * v2)
Next, let's solve for v2:
0.145 kg * v2 - 0.025 kg * v2 = 0.0672 kg m/s
0.12 kg * v2 = 0.0672 kg m/s
v2 = 0.0672 kg m/s / 0.12 kg
v2 = 0.56 m/s / 4
v2 = 0.14 m/s
Therefore, the change in speed of the car is:
Δv = v2 - v1 = 0.14 m/s - 0.56 m/s = -0.42 m/s
The negative sign indicates that the car's speed has decreased by 0.42 m/s.