A 2.40 kg mess kit sliding on a frictionless surface explodes into two 1.20 kg parts:
3.78 m/s, due north, and 5.28 m/s, 32.8° north of east. What is the original speed of the mess kit?
M1*V1 = M2*V2 + M3*V3
2.4*V1 = 1.2*3.78[90o] + 1.2*5.28[32.8o]
2.4V1 = 4.54i + 4.44 + 5.33 + 3.43i
2.4V1 = 9.77 + 7.97i
V1 = 4.07 + 3.32i
V1 = sqrt(4.07^2+3.32^2) = 5.25 m/s
To find the original speed of the mess kit, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion is equal to the total momentum after the explosion.
Initially, the mess kit is a single object, so its momentum can be calculated as the product of its mass and velocity. Let's denote the original velocity of the mess kit as v.
The momentum before the explosion is the sum of the momenta of the two parts after the explosion. Let's denote the velocity of the first part as v1 (3.78 m/s, due north) and the velocity of the second part as v2 (5.28 m/s, 32.8° north of east).
The momentum of the first part is given by the product of its mass (1.20 kg) and velocity (v1).
The momentum of the second part is given by the product of its mass (1.20 kg) and velocity (v2).
So, according to the conservation of momentum, we have:
(initial momentum) = (momentum of the first part) + (momentum of the second part)
mv = m1v1 + m2v2
Plugging in the given values, we have:
2.40 kg * v = 1.20 kg * (3.78 m/s) + 1.20 kg * (5.28 m/s * cos(32.8°))
Now, we can solve this equation to find the value of v. Let's do the calculations:
2.40 kg * v = 1.20 kg * 3.78 m/s + 1.20 kg * (5.28 m/s * cos(32.8°))
2.40 kg * v = 3.78 kg m/s + 1.20 kg * (5.28 m/s * 0.848)
2.40 kg * v = 3.78 kg m/s + 1.20 kg * 4.464 m/s
2.40 kg * v = 3.78 kg m/s + 5.357 kg m/s
2.40 kg * v = 9.137 kg m/s
v = 9.137 kg m/s / 2.40 kg
v = 3.807 m/s (rounded to three decimal places)
Therefore, the original speed of the mess kit was approximately 3.807 m/s.