a 4.0 kg object has a velocity of 4.0 m/s headed east. an explosion causes it to break up into two smaller 2.0 kg objects. if one of the objects has a velocity of 3.0 m/s at an angle of 30 degrees east of north, what is the velocity of the other object after the explosion

To find the velocity of the other object after the explosion, we can first determine the velocity of the system before the explosion and then use the law of conservation of momentum.

The momentum of an object is given by the product of its mass and velocity. So, let's calculate the initial momentum of the 4.0 kg object before the explosion.

Initial momentum of the 4.0 kg object = mass × velocity
= 4.0 kg × 4.0 m/s (headed east)

Since momentum is a vector quantity, we need to consider the direction as well. Since the object is moving east, the momentum is in the eastward direction.

Momentum of the 4.0 kg object before the explosion = 16 kg⋅m/s (east)

According to the law of conservation of momentum, the total momentum before the explosion is equal to the total momentum after the explosion. Since the objects break up into two smaller 2.0 kg objects, the total mass after the explosion is still 4.0 kg.

Now, let's assume that the other object has a velocity of v2 m/s after the explosion. The other object's mass and velocity will be 2.0 kg and v2 m/s, respectively.

Applying the law of conservation of momentum, we can write:

Initial momentum of the system before the explosion = Total momentum of the objects after the explosion

Initial momentum of the system before the explosion = (mass1 × velocity1) + (mass2 × velocity2)

16 kg⋅m/s (east) = (2.0 kg × 3.0 m/s) + (2.0 kg × v2)

16 kg⋅m/s (east) = 6 kg⋅m/s (east of north) + 2 kg⋅m/s (v2)

We need to consider the direction as well. The velocity of one object is given as 3.0 m/s at an angle of 30 degrees east of north. So, the momentum contribution of this object is 6 kg⋅m/s (east of north) based on its magnitude and direction.

Substituting the values, we can solve for v2:

16 kg⋅m/s (east) = 6 kg⋅m/s (east of north) + 2 kg⋅m/s (v2)

16 kg⋅m/s (east) - 6 kg⋅m/s (east of north) = 2 kg⋅m/s (v2)

Now, to calculate the magnitude and direction of v2, we can use trigonometry. The magnitude of v2 can be found using the Pythagorean theorem:

|v2| = sqrt[ (16 kg⋅m/s - 6 kg⋅m/s)^2 + (2 kg⋅m/s)^2]

|v2| = sqrt[ (10 kg⋅m/s)^2 + (2 kg⋅m/s)^2]

|v2| = sqrt[100 kg^2⋅m^2/s^2 + 4 kg^2⋅m^2/s^2]
|v2| = sqrt[104 kg^2⋅m^2/s^2]
|v2| ≈ 10.20 m/s

The direction of v2 can be found by using trigonometry. Since we have two vectors, v2 and 6 kg⋅m/s (east of north), we can use the inverse tangent (arctan) function to find the angle.

tan(angle) = (opposite / adjacent)
tan(angle) = (2 kg⋅m/s / 6 kg⋅m/s)

angle = arctan(2/6)
angle ≈ 18.4 degrees

Therefore, the magnitude and direction of the velocity of the other object after the explosion are approximately 10.20 m/s at an angle of 18.4 degrees east of north.