A sled that has a mass of 8.8kg moves horizontally at 20 km/h. When it passes under a bridge 13kg of snow drops ontothe sled. What is its subsequent speed?
Isn't this in inellastic collision equation. I used (m1v1)+(m2v2)=(m1+m2)(vf)
m1=8.8kg
v1=20km/h
m2=21.8kg
v2=0km/h
I got 5.75 km/h where did i go wrong?
To solve this problem using the inelastic collision equation, you correctly identified the initial values:
m1 = 8.8 kg (mass of the sled)
v1 = 20 km/h (initial velocity of the sled)
m2 = 13 kg (mass of the snow)
v2 = 0 km/h (initial velocity of the snow)
However, you made a mistake while calculating the final velocity. Let's correct it:
Using the equation: (m1 * v1) + (m2 * v2) = (m1 + m2) * vf
Now, let's plug in the values:
(8.8 kg * 20 km/h) + (13 kg * 0 km/h) = (8.8 kg + 13 kg) * vf
(176 kg·km/h) + 0 kg·km/h = 21.8 kg * vf
176 kg·km/h = 21.8 kg * vf
To find the final velocity, divide both sides by 21.8 kg:
vf = (176 kg·km/h) / 21.8 kg
vf ≈ 8.077 km/h (rounded to three decimal places)
Therefore, the subsequent speed of the sled, after the 13 kg of snow drops onto it, is approximately 8.077 km/h.