A bullet of mass 100g is fired from a gun of mass 20kg with a velocity of 100m/s. Calculate the recoil velocity of the gun.
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let, the mass of bullet is m and mass of the gun is M.
the speed of the bullet v and speed of gun is V
then according to conservation of momentum
MV + mv=0
here M= 20kg
m=100g= 0.1 kg
V= ?
v=100 m/s
so,
V= -mv÷M
= - 0.5m/s
{ the (-) sign denote the Gun's velocity is opposite to bullet's velocity}
the speed of the bullet v and speed of gun is V
then according to conservation of momentum
MV + mv=0
here M= 20kg
m=100g= 0.1 kg
V= ?
v=100 m/s
so,
V= -mv÷M
= - 0.5m/s
{ the (-) sign denote the Gun's velocity is opposite to bullet's velocity}
afifadawre:
Hello
Can you please help me with this question too?
An Iron sphere of mass 10kg is dropped from a height of 80cm. If the downward acceleration of the ball is 10m/s^2, calculate the momentum transferred to the ground by the ball.
When a body fall from height h the velocity at the moment when it touch the ground it's velocity is
V=√(2gh)
Here g=10m/s^2
h=80cm=0.8m
So the momentum of the body when it touch the ground is
p=m×v
=10×√(2×10×0.8)
=10×4
=40 kg m/ sec
As after touch the ground the body is at rest so total momentum is transfered to ground.
So, momentum transfer to ground is
P= 40kg m/sec
V=√(2gh)
Here g=10m/s^2
h=80cm=0.8m
So the momentum of the body when it touch the ground is
p=m×v
=10×√(2×10×0.8)
=10×4
=40 kg m/ sec
As after touch the ground the body is at rest so total momentum is transfered to ground.
So, momentum transfer to ground is
P= 40kg m/sec
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