In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired
1 answer:
The velocity of the bullet and the block just after impact is mathematically given as
u=6.647m/s
v=1.534m/s
<h3> Velocity of the bullet and the block</h3>Question Parameters:
In the ballistic pendulum experiment, a bullet of mass 0.06 kg is fired
horizontally into a wooden block of mass 0.2 kg.
the block is 0.12 m above its initial position.
Generally the equation for the conservation of energy is mathematically given as
0.5(m+n)v^2=(m+n)gh
Therefore
v^2=2gh
v=1.534m/s
Hence
mu+m*0=(m+n)v
0.060*u=(0.060+0.2)*1.534
u=6.647m/s
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C. The downward component of the projectile's velocity continually increases
Explanation:
The motion of a projectile consists of two independent motions:
- A uniform motion (with constant velocity) along the horizontal direction
- A uniformly accelerated motion, with constant acceleration (equal to the acceleration of gravity) in the downward direction
Here we want to study the downward component of the projectile's velocity. Since the vertical motion is a uniformly accelerated motion, the vertical velocity is given by:
where
u = 0 is the initial vertical velocity (zero since the projectile is fired horizontally)
downward is the acceleration of gravity
t is the time
So the equation becomes
This means that
C. The downward component of the projectile's velocity continually increases
Because every second, it increases by in the downward direction.
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