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schepotkina [342]
4 years ago
14

Find the speed vfinal of the joined cars after the collision. mastering physics

Physics
1 answer:
Tanya [424]4 years ago
6 0
<span>Px = 0 Py = 2mV second, Px = mVcosφ Py = –mVsinφ add the components Rx = mVcosφ Ry = 2mV – mVsinφ Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²) and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) simplifying Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²) Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²) Vf = (V/3)âš((cosφ)² + (2 – sinφ)²) Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ)) Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ)) using the identity sin²(Ď)+cos²(Ď) = 1 Vf = (V/3)âš1 + 4 – 2sinφ) Vf = (V/3)âš(5 – 2sinφ)</span>
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Answer: 0.642mm

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We will assume the motion of the object to be of a constant acceleration, hence newton's laws of motion is applicable.

Recall that f = ma.

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This acceleration of vertical because it occurred when the object deflected.

5.2×10^-16 = 9.11×10^-31 (ay)

ay = 5.2×10^-16 / 9.11×10^-31

ay = 5.71×10^14 m/s²

For the horizontal motion, x = vt

Where x = horizontal distance = 0.019m and v is the velocity = 1.2×10^7 m/s,

By substituting the parameters, we have that

0.019 = 1.27×10^7 × t

t = 0.019 / 1.27 × 10^7

t = 1.5×10^-9 s

The vertical distance (y) is gotten by using the formulae below

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Answer:

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