1)Particle A and particle B are held together with a compressed spring between t

Question

1)Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes them apart and they then fly off in opposite directions, free of the spring. The mass of A is 5.00 times the mass of B, and the energy stored in the spring was 80 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particleA and (b) particle B?


2)The magnitude of an unbalanced force on a 5.9 kg object increases at a constant rate from zero to 61 N in 2.2 s, causing the initially stationary object to move. What is the object's speed at the end of the 2.2 s?

Explanation / Answer

1)

mass of the particelA is m1 and speed is v1

mass of the particelB is m2 and speed is v2

given m1=5*m2

petential enegry stored, U=80 J

use law of conservation of momentum,

m1*v1+m2*v2=0

5*m2*v1+m2*v2=0

====> v2=-5*v1

K.E of A is, K1=1/2*m1*v1^2

K.E of B is, K2=1/2*m2*v2^2

===> K2=1/2*(m1/5)*(-5*v1)^2

K2=1/2*5*m1*v1^2

=====> K2=5*k1

now,

by using law of conservation of enegry,

U=K1+K2

80=K1+5*K1

===> K1=13.33 J

and

K2=5*K1

===> K2=66.66 J

a)

kinetic enegry of particle A is K1=13.33 J

b)

kinetic enegry of particle B is K2=66.66 J

2)

mass of the object, m=5.9kg

force F1=0 and F2=61 N

time t=2.2 sec

use,

avg force, Favg=(61+0)/2 =30.5 N

avg acceleration, a=Favg/m

a=30.5/5.9

a=5.17 m/sec^2

after t=2.2 sec,

speed v=u+a*t

v=0+5.17*2.2

v=11.374 m/sec

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