An object with total mass m total = 17 kg is sitting at rest when it explodes in

Question

An object with total mass mtotal = 17 kg is sitting at rest when it explodes into three pieces. One piece with mass m1 = 4.9 kg moves up and to the left at an angle of 1 = 24° above the –x axis with a speed of v1 = 28.5 m/s. A second piece with mass m2 = 5.4 kg moves down and to the right an angle of 2 = 29° to the right of the -y axis at a speed of v2 = 23.7 m/s.

1) What is the magnitude of the final momentum of the system (all three pieces)?

2) What is the mass of the third piece?

3) What is the x-component of the velocity of the third piece?

4) What is the y-component of the velocity of the third piece?

5) What is the magnitude of the velocity of the center of mass of the pieces after the collision?

6) Calculate the increase in kinetic energy of the pieces during the explosion.

Explanation / Answer

1) This is a conservation of momentum problem. since the explosion represents an internal force, there is no change in momentum, and the magnitude of the system's momentum remains zero.

2. mass is conserved, so the sum of the three masses must be 17kg, thus

m3 = 17kg - 4.8 kg - 5.4 kg = 6.8 kg

3. The sum of all x components of momentum must be zero; the first two fragments have x momenta of
=> Vx = Vcos(24)
=> 4.9kg * (-28.5 cos 24) m/s = -127.57 kg-m/s

=> 5.4 kg * 23.7 sin 29 m/s = 62.04 kg-m/s

therefore, the x component of the third fragment is -(62.04-127.57) = +65.53kg m/s

so its velocity = 65.53/ 6.8 = 9.636 m/s

the y components of momentum are

=> 4.9kg * 28.5 sin 24 - 5.4 kg * 23.7 cos 29 =56.79 - 111.93 = -55.14kg-m/s

so the y component of the third fragment is + 55.14 kgm/s

and its velocity in the y direction = 55.14/6.8 = 8.1 m/s

5) System momentum still = 0, so velocity of the center of mass = 0

6) velocity of third mass is V = sqrt( 9.636² + 8.1²) m/s = 12.59 m/s
=> increase in kinetic energy

Ek = 1/2 * 4.9kg * (28.5m/s)² + 1/2 * 5.4kg * (23.7m/s)² + 1/2 * 6.8kg * (12.59m/s)² = 4045.5 J

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