a) 6.1 m/s
b) 19.0 m
Explanation:
a)
Before the collision, the initial gravitational potential energy of the pendulum is converted into kinetic energy. So, we can write:
[tex]m_1gh=\frac{1}{2}m_1u^2[/tex]
where
[tex]m_1[/tex] is the mass of the steel ball attached to the pendulum
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
h = 1.9 m is the initial height of the steel ball (since the pendulum is in horizontal position)
u is the velocity of the steel ball just before hitting the block
Solving for u,
[tex]u=\sqrt{2gh}=\sqrt{2(9.8)(1.9)}=6.1 m/s[/tex]
After that, the steel ball collides elastically with the block, of mass
[tex]m_2=1.1 kg[/tex]
The collision is elastic, this means that both total momentum and total kinetic energy are conserved, so we can write:
[tex]m_1 u = m_1 v_1 + m_2 v_2\\\frac{1}{2}m_1 u^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2[/tex]
where
[tex]v_1[/tex] is the final velocity of the steel ball
[tex]v_2[/tex] is the final velocity of the block
Solving simultaneously the two equations, we can find the final velocity of the block:
[tex]v_2=\frac{2m_1}{m_1+m_2}u=\frac{2(1.1)}{1.1+1.1}(6.1)=6.1 m/s[/tex]
Because the two objects have same mass, so the steel ball has transferred all its energy and momentum to the block.
b)
After the collision, the block slides with a certain deceleration given by the presence of the force of friction.
The force of friction on the block is
[tex]F_f=-\mu mg[/tex]
where
[tex]\mu=0.10[/tex] is the coefficient of friction
m = 1.1 kg is the mass of the block
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
The acceleration of the block is therefore (Newton's second law):
[tex]a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g[/tex]
Since the motion is a uniformly accelerated motion, we can use the suvat equation to determine the distance the block covers before coming to rest:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity
u = 6.1 m/s is the initial velocity
[tex]a=-\mu g[/tex] is the acceleration
s is the stopping distance
Solving for s,
[tex]s=\frac{v^2-u^2}{2(-\mu g)}=\frac{0^2-(6.1)^2}{2(-0.10)(9.8)}=19.0 m[/tex]
