A 1000-kg car approaches an intersection traveling north at 20.0 m/s. A 1200-kg car approaches the same intersection traveling e ast at 22.0 m/s. The two cars collide at the intersection and lock together. Ignoring any external forces that act on the cars during the collision, what is the velocity of the cars immediately after the collision?
2 answers:
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:
During the collision of two cars there is no external force on this system
so the total momentum of two cars will remains conserved
so we will have
now when two cars collide inelastically then they will move together after collision
so we have
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Answer:
θ=142.9°
Explanation:
d=1 *r
angle ϕ= 37.1°
the line connecting pebble and target should be tangent to a circle so
cos(180-ϕ-θ)= =
∴ θ=180-ϕ-
θ= 180-37.1-0
θ=142.9°
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