Physics, 24.03.2020 01:13 kalcloud

A steel ball of mass m1 = 1.1 kg and a cord of length of L = 1.9 m of negligible mass make up a simple pendulum that can pivot without friction about the point O, as in the figure below. This pendulum is released from rest in a horizontal position, and when the ball is at its lowest point it strikes a block of mass m2 = 1.1 kg sitting at rest on a shelf. Assume that the collision is perfectly elastic and that the coefficient of kinetic friction between the block and shelf is 0.10.(a) What is the velocity of the block just after impact?(b) How far does the block slide before coming to rest (assuming that the shelf is long enough)?

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Answer from: sean7900

a) 6.1 m/s

b) 19.0 m

Explanation:

Before the collision, the initial gravitational potential energy of the pendulum is converted into kinetic energy. So, we can write:

$m_1gh=\frac{1}{2}m_1u^2$

where

m_1 is the mass of the steel ball attached to the pendulum

g=9.8 m/s^2 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,

$u=\sqrt{2gh}=\sqrt{2(9.8)(1.9)}=6.1 m/s$

After that, the steel ball collides elastically with the block, of mass

m_2=1.1 kg

The collision is elastic, this means that both total momentum and total kinetic energy are conserved, so we can write:

$m_1 u = m_1 v_1 + m_2 v_2\\\frac{1}{2}m_1 u^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

where

v_1 is the final velocity of the steel ball

v_2 is the final velocity of the block

Solving simultaneously the two equations, we can find the final velocity of the block:

$v_2=\frac{2m_1}{m_1+m_2}u=\frac{2(1.1)}{1.1+1.1}(6.1)=6.1 m/s$

Because the two objects have same mass, so the steel ball has transferred all its energy and momentum to the block.

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

$F_f=-\mu mg$

where

$\mu=0.10$ is the coefficient of friction

m = 1.1 kg is the mass of the block

g=9.8 m/s^2 is the acceleration due to gravity

The acceleration of the block is therefore (Newton's second law):

$a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g$

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:

v^2-u^2=2as

where

v = 0 is the final velocity

u = 6.1 m/s is the initial velocity

$a=-\mu g$ is the acceleration

s is the stopping distance

Solving for s,

$s=\frac{v^2-u^2}{2(-\mu g)}=\frac{0^2-(6.1)^2}{2(-0.10)(9.8)}=19.0 m$

Answer from: Quest

Answerwavelength;

Answer from: Quest

note: the given question is incomplete.

complete question is given below.

which is an example of kinetic energy?

a. the energy stored in ethanol

b. a hockey puck sliding across the ice

c. a compressed spring

d. a ball sitting at the top of a ramp

solution: the kinetic energy is the energy posses by the virtue of the motion of the object.

in the given options, a, c and d are the example of potential energy because it is the energy posses by the virtue of the rest of the object. but option b has the kinetic energy as the object is in motion.

hence, a hockey puck sliding across the ice is the apex example of kinetic energy.

so, the option (b) a hockey puck sliding across the ice is the correct option.

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A steel ball of mass m1 = 1.1 kg and a cord of length of L = 1.9 m of negligible mass make up a simp...