Answer:
v = 1.28 m/s
Explanation:
Given that,
Maximum compression of the spring, 
Spring constant, k = 800 N/m
Mass of the block, m = 0.2 kg
To find,
The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.
Solution,
When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,
Initial energy = Final energy
Substituting all the values in above equation,
v = 1.28 m/s
Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.
Answer:
on an average, <em>2</em><em>0</em><em>H</em><em>z</em><em> </em><em>to</em><em> </em><em>2</em><em>0</em><em>k</em><em>H</em><em>z</em>
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
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