physics help please ?

Question

part 1 A bullet of mass 2 g moving with an initial
speed 500 m/s is fired into and passes through
a block of mass 6 kg, as shown in the figure.
The block, initially at rest on a frictionless,
horizontal surface, is connected to a spring of
force constant 599 N/m. If the block moves a distance 0.68 cm to the
right after the bullet passed through it, find
the speed v at which the bullet emerges from
the block.
Answer in units of m/s.

part 2 Find the magnitude of the energy lost in the
collision.
Answer in units of J.

part 1 A 580 kg mass is sliding on a horizontal frictionless surface with a speed of 9.5 m/s when
it collides with a 82 kg mass initially at rest,
as shown. The masses stick together and slide
up a frictionless track at 60◦
from horizontalWhat is the speed of the two blocks immediately after the collision? The acceleration
due to gravity is 9.8 m/s
2
.
Answer in units of m/s.

part 2 After the collision, the masses slide up the
incline. To what maximum height h above the horizontal surface will the masses slide?
Answer in units of m

thank you for the help whoever does this have a blessed day

Whome · Accepted Answer

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1a) The kinetic energy of the block converts to spring potential
ASSUME the block does not lose mass as the bullet rips through it.

½mv² = ½kx²
v = √(kx²/m) = √(599(0.0068)²/ 6 = 0.0679 m/s

Conservation of momentum during the collision.

0.002(500) + 6(0) = 0.002v + 6(0.0679)
v = 296.17... ≈ 300 m/s   ANSWER

1b) ΔKE = ½(0.002)500² - ½(0.002)296² - ½(6)0.0679² 
ΔKE = 162.269... ≈ 160 J  ANSWER

2a)  Conservation of momentum

580(9.5) + 82(0) = (580 + 82)v
v = 8.32326... ≈ 8.3 m/s  ANSWER

2b) Conservation of energy

PE = KE
mgh = ½mv²
     h = v²/2g
     h = 8.32326²/2(9.8) = 3.534... ≈ 3.5 m  ANSWER

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physics help please ?

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