2.What is the speed of the larger piece after the collision. 3.if the explosion
ID: 1782851 • Letter: 2
Question
2.What is the speed of the larger piece after the collision.
3.if the explosion lasted for a time t=.027s what was the average force on the larger piece
4.what is the magnitude of the charge in momentum of the smaller piece
5. what is the magnitude of the velocity of the center of mass of pieces after the collision.
L, Fliplt Physics-20174 F × Homework: 2130 Exam- A Racquet Ball With Mas × u Math ay Math Proble × cià Secure! https://www.flipitphys unitltemlD 3242698BenrollmentiD-263475 Standard Exercise Car-Truck Cc Explosion 2 1 2 3 4 5 6 Standard Exercise Ball Hits Wal An object with total mass mtotal 6.1 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 43 J Standard Exercise Bumper Cars 1) What is the speed of the smaller piece after the collision? Standard Exercise Three Masse m/s Submit You currently have O submissions for this question. Only 2 submission are allowed You can make 2 more submissions for this question Standard Exercise System of Pa 2) What is the speed of the larger piece after the collision? m/s Submit Standard Exercise Two Train Ca You currently have 0 submissions for this question. Only 2 submission are allowed You can make 2 more submissions for this question Standard Exercise Explosion 2 3) If the explosion lasted for a time t = 0.027 s, what was the average force on the larger piece? N Submit You currently have 0 submissions for this question. Only 2 submission are allowed You can make 2 more submissions for this question Standard Exercise Tipler6 9.P.C 9:23 PM 1115/2017 Type here to searchExplanation / Answer
he momentum before the explosion = 0 so the momentum after the explosion is also = 0
mv + 3mV = 0
By the energy given
0.5mv^2 + 0.5(3m)V^2 = 43
solving for V in the momentum equation
V = -mv/3m = -v/3
Plugging back into the energy equation
0.5mv^2 + 0.5(3m)(-v/3)^2 = 43
0.5mv^2 + (1.5/9)mv^2 = 43
v^2 = 43/(2/3)m
4m = 6.1 kg so m = 1.525 kg
v^2 = 64.5/1.525
v = 6.5 m/s =velocity of smaller piece
2) V = 3v = 3(6.5) = 19.51 m/s=velocity of larger piece
3)Ft = (3m)V
F(0.027) = 3(1.525)(19.51)
F = 3305.86 N
4)change in momentum for smaller> 1.525(6.5) = 9.91 kg m/s
5) velocity of the center of mass = 0
Again by conservation of momentum
momentum of the system is always 0 the momentum for a system of particles is the total mass of the system
times the velocity of the center of mass, since there is mass velocity of center of mass = 0 m/s
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