This problem applies momentum balance over the cars collided
m1v1y + m2v2y = (m1 + m2)v'y
<span>0 + (1500 kg)(25.0 m/s) = (2500 kg)v'y </span>
<span>v'y = 15.0 m/s </span>
<span>Find the vector sum of the velocities: </span>
<span>v^2 = v'x^2 + v'y^2 = (8.00 m/s)^2 + (15.0 m/s)^2 </span>
<span>v = 17.0 m/s </span>
<span>find the angle: θ = tan^-1(15.0 m/s / 8.00 m/s) = 61.9° </span>
<span>now, the momentum: p = mv = (2500 kg)(17.0 m/s) </span>
<span>p = 4.25 x 10^4 kg x m/s at 61.9° N of E</span>
Answer:
Explanation:
Ignoring friction and assuming one ramp end rests on level ground.
gravity acceleration acting parallel to the ramp is gsinθ.
F = mgsinθ
mg = 250 N, sinθ = 3/25
F = 250(3/25)
F = 30 N
A 25 meter long ramp strong enough to hold a person plus a 250 N refrigerator would weigh much more than 250 N.
0 m.s because once it came to the ground it’s stopped