An object with total mass m total = 6.3 kg is sitting at rest when it explodes i

Question

An object with total mass mtotal = 6.3 kg is sitting at rest when it explodes into two pieces. The two pieces, after the explosion, have masses of m and 3m. During the explosion, the pieces are given a total energy of E = 44 J.

1)What is the speed of the smaller piece after the collision? (m/s)

2)What is the speed of the larger piece after the collision? (m/s)

3)If the explosion lasted for a time t = 0.028 s, what was the average force on the larger piece? (N)

4)What is the magnitude of the change in momentum of the smaller piece? (kg-m/s)

5)What is the magnitude of the velocity of the center of mass of the pieces after the collision? (m/s)

Explanation / Answer

Mtotal = 6.3 kg   ,

V = 0   = initial velocity of object

m1 = m

m2 = 3 m

m1 + m2 = 6.3 kg

m + 3 m = 6.3 kg

m = 1.575 kg

m1 = 1.575

m2 = 3 m = 4.725 kg

using conservation of momentum ::

mV = m1 V1 + m2 V2
m (0) = m V1 + 3m V2

V1 = - 3 V2

Using conservation of energy ::

KE of m1 + KE of m2 = 44

(0.5) m1 V12 + (0.5) m2 V22 = 44

m (-3V2)2 + (3m) (V2)2 = 88

6 m V22 = 44

6 (1.575) V22 = 44

V2 = 2.16 m/s
v1 = - 3 V2 = -3 x 2.16 = -6.48 m/s

1) speed of smaller block = V1 = - 6.48 m/s

2) speed of larger block = V2 = 2.16 m/s

3) Using the formula , impulse = change in momentum

F x t = change in momentum

t = 0.028 sec

change in momentum of larger piece is given as ::

final momentum - initial momentum = (m2) (V2) - 0 = 4.725 x 2.16 = 10.21 kg m/s

F (0.028) = 10.21

F = 364.6 N

4) change in momentum of smaller mass = m1 V1 -0 = 1.575 x 6.48 = 10.21 kgm/s

5) velocity of center of mass = 0 m/s

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