Question

An object of 7kg is lit and it becomes suspended at rest from the ceiling. It then explodes into 3 parts. Part 1 has a mass of 2.7kg and is traveling 3.1m/s at 22 degrees above x-axis. Part 2 has a mass of 1.75kg and moves 4.4 m/s at 53 degrees above x-axis. Determine the mass, velocity, and direction of the 3rd piece using momentum. Thank you.

Answer 1

Mass of first part, m1 = 2.7 kg

Mass of second part, m2 = 1.75 kg

So, mass of the third part, m3 = 7 kg - (2.7 kg + 1.75 kg)

= 2.55 kg (Answer)

Write the component of velocity in unit vector notations.

Velocity of first part, v1 = [(3.1*cos22)i + (3.1*sin22)j] m/s

= [2.87i + 1.16j] m/s

Velocity of second part, v2 = [(4.4*cos53)i + (4.4*sin53)j] m/s

= [2.65i + 3.51j] m/s

Suppose the velocity of the third part is v3.

Now, initial momentum of the combined object, Pi = 0

Final momentum of the three parts -

Pf = m1*v1 + m2*v2 + m3*v3

= 2.7 kg*[2.87i + 1.16j] m/s + 1.75 kg*[2.65i + 3.51j] m/s + 2.55 kg *v3

According to conservation of momentum -

Pi = Pf

=> 0 = 2.7 kg*[2.87i + 1.16j] m/s + 1.75 kg*[2.65i + 3.51j] m/s + 2.55 kg *v3

=> 0 = (12.39 kg*m/s)i + (9.27 kg*m/s)j + 2.55 kg *v3

=> 2.55 kg *v3 = -[(12.39 kg*m/s)i + (9.27 kg*m/s)j]

=> v3 = -[4.86i + 3.64j] m/s

Therefore, magnitude of the third velocity -

|v3| = sqrt[(4.86)^2 + (3.64)^2] = 6.07 m/s (Answer)

Direction of the velocity, = 180 + tan^-1(3.64/4.86)

= 180 + 36.8 = 216.8 deg from the positive direction of x-axis.