An object with total mass m total = 6.3 kg is sitting at rest when it explodes i
ID: 1376343 • Letter: A
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