An eruption of energy unfolds, embodying chaos and change. Shards are thrown arbitrarily, one large fragment races along the path of the positive x-axis, its momentum creating a wake of disturbance. Simultaneously, another fragment, massive and unpredictable, disobeys gravity as it hurtles in a diagonal trajectory towards the negative x-axis and negative y-axis. The sheer force of their movements leaves a disarray of wreckage and swirling detritus in their wake; evidence of the intense spectacle that has just occurred.

A 10 kg object with a velocity 7.5i m/s explodes into two equal fragments. One flies off with a velocity 2.5i -2.5j m/s.

What is the velocity of the other fragment in vector notation?

initial momentum = mV = 10(7.5 i + 0 j)

= 75 i + 0 j

final momentum = 5(2.5 i -2.5 j)+ 5(Vxi+Vyj)
= (12.5 +5Vx) i (-12.5 +5Vy)j

final momentum = initial momentum
x direction:
12.5 + 5 Vx = 75
Vx = 12.5

y direction
-12.5 + 5 Vy = 0
Vy = 2.5

so V = 12.5 i + 2.5 J

thank youuuu so much for the explanation :D

You are welcome :)

Amazing simplicity. You have a talent

Well, it seems like there was quite an explosive breakup! To find the velocity of the other fragment, we can use the principle of conservation of momentum.

Since the initial momentum of the 10 kg object is given by momentum_initial = mass * velocity_initial, we can calculate its initial momentum as momentum_initial = 10 kg * 7.5i m/s.

Now, let's denote the velocity of the other fragment as v2. According to the law of conservation of momentum, the total momentum before the explosion should be equal to the total momentum after the explosion. So, we have:

momentum_initial = momentum_fragment1 + momentum_fragment2

Substituting the given values, we get:

10 kg * 7.5i m/s = 10 kg * (2.5i - 2.5j) m/s + 10 kg * v2

Simplifying the equation, we have:

75i kg m/s = 25i kg m/s - 25j kg m/s + 10 kg * v2

Now, let's isolate the velocity of the other fragment, v2:

10 kg * v2 = 75i kg m/s - 25i kg m/s + 25j kg m/s

v2 = (75i kg m/s - 25i kg m/s + 25j kg m/s) / 10 kg

v2 = 5i kg m/s + 2.5j kg m/s

So, the velocity of the other fragment, in vector notation, is 5i m/s + 2.5j m/s. Just be careful, it may be a bit shaken up after the explosive breakup!

To find the velocity of the other fragment, we can use the principle of conservation of momentum. According to this principle, the total momentum before the explosion should be equal to the total momentum after the explosion.

The total momentum before the explosion can be calculated by multiplying the mass of the object by its initial velocity:

Initial momentum = (mass) x (initial velocity)

Given that the mass of the object is 10 kg and its initial velocity is 7.5i m/s, the initial momentum is:

Initial momentum = 10 kg x 7.5i m/s

Now, let's consider the momentum after the explosion. We know that one fragment flies off with a velocity of 2.5i - 2.5j m/s. To find the velocity of the other fragment (let's call it v):

Final momentum = (mass of fragment 1) x (velocity of fragment 1) + (mass of fragment 2) x (velocity of fragment 2)

Since the fragments are equal in mass, we can assume that they have the same mass. Therefore, the final momentum can be written as:

Final momentum = 2 x (mass) x (velocity of one fragment) + (mass) x (v)

Given that the final momentum is equal to the initial momentum, we can equate the equations for the initial and final momentum:

10 kg x 7.5i m/s = 2 x (mass) x (2.5i - 2.5j m/s) + (mass) x (v)

Simplifying the equation, we get:

75i kg·m/s = 5i kg·m/s - 5j kg·m/s + (mass) x (v)

Since we know the velocity of one fragment, we can isolate the velocity of the other fragment (v) by equating the coefficients of the i and j terms:

75i = 5i + (mass) x (v) --> 70i = (mass) x (v)

Now, we substitute the mass of the object (10 kg) into the equation:

70i = 10 kg x v

Simplifying further, we can find the velocity of the other fragment in vector notation:

v = 7i m/s

Therefore, the velocity of the other fragment is 7i m/s.