A sledthat 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=20km/h
I got 20km/s where did i go wrong?
V2 is zero. The snow was not moving horizontally when if fell.
To solve this question, we need to use the principle of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.
The initial momentum of the sled can be calculated as (m1 * v1), where m1 is the mass of the sled (8.8 kg) and v1 is its initial velocity (20 km/h).
m1 * v1 = (8.8 kg) * (20 km/h)
To determine the final velocity of the combined system (sled + snow), we need to calculate the total mass of the system and apply the conservation of momentum equation:
Total mass = m1 + m2 = 8.8 kg + 13 kg
Then, the final velocity (vf) can be found by dividing the initial momentum by the total mass:
vf = (m1 * v1) / (m1 + m2)
Plugging in the values:
vf = [(8.8 kg) * (20 km/h)] / (8.8 kg + 13 kg)
Now, let's perform the calculation:
Total mass = 8.8 kg + 13 kg = 21.8 kg
vf = (8.8 kg * 20 km/h) / (21.8 kg)
Calculating this expression, we get:
vf ≈ 8.07 km/h
Therefore, the subsequent speed of the sled, after the snow drops onto it, is approximately 8.07 km/h.