Since the bag was at rest, its initial momentum is zero. The velocity of the ball before collision is 500 ms-1.
<h3>Linear momentum</h3>
The term momentum in physics refers the product of mass and velocity. If we know mass of the object and its velocity, then we calculate the momentum.
Momentum before collision for the bullet = 0.01 kg × v
Momentum before collision for the bag = 0
Momentum after collision for the bag and bullet = (0.01 kg + 0.49 kg) 10 = 5 Kgms-1
The velocity of the bullet before collision = 0.01 kg × v + 0 = 5 Kgms-1
v = 5 Kgms-1/0.01 kg
v = 500 ms-1
Learn more about momentum: brainly.com/question/904448
Answer:
speed of the bullet before it hit the block is 200 m/s
Explanation:
given data
mass of block m1 = 1.2 kg
mass of bullet m2 = 50 gram = 0.05 kg
combine speed V= 8.0 m/s
to find out
speed of the bullet before it hit the block
solution
we will apply here conservation of momentum that is
m1 × v1 + m2 × v2 = M × V .............1
here m1 is mass of block and m2 is mass of bullet and v1 is initial speed of block i.e 0 and v2 is initial speed of bullet and M is combine mass of block and bullet and V is combine speed of block and bullet
put all value in equation 1
m1 × v1 + m2 × v2 = M × V
1.2 × 0 + 0.05 × v2 = ( 1.2 + 0.05 ) × 8
solve it we get
v2 = 200 m/s
so speed of the bullet before it hit the block is 200 m/s
If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.
It is also tripled, there is a rule to everything, whatever you do to one thing, you do the exact thing to the other. Hope this solves it :)
