Physics, 29.07.2019 04:00 jaylaa04

A15 kg block is attached to a very light horizontal spring of spring constant 500 n/m and is resting on a frictionless horizontal table as shown in the figure. suddenly it is struck by a 3 kg stone traveling horizontally at 8 m/s to the right; the stone rebounds at 2 m/s horizontally to the left. find the maximum distance that the block will compress the spring after the collision.

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Let's separate the problem in two parts:

Part 1): Collision between the stone and the block.
In the collision, the total momentum of the system stone+block is conserved.
Before the collision, only the stone is moving, so the total momentum is:
p_i = m_s v_s

where

is the mass of the stone and v_s = 8 m/s

is the speed of the stone, traveling towards the block (to the right).
After the collision, both the stone and the block are in motion, so the total momentum is:
p_f = m_s v_s'+m_b v_b

where

is the new speed of the stone (with a negative sign, since the stone is now moving in the opposite direction, to the left), m_b = 15 kg

is the mass of the block and v_b

is the mass of the block just after the collision.
Since the momentum must be conserved,
p_i = p_f

So we can rewrite everything
m_s v_s = m_s v_s' + m_b v_b

to find

$v_b = \frac{m_s v_s - m_s v_s'}{m_b}= \frac{(3kg)(8m/s)-(3kg)(-2m/s)}{15kg}=2 m/s$
So the block slides with speed 2 m/s to the right after the collision.

Part 2) Block compressing the spring
At this point we can ignore the stone and focus only on the block and the spring. The block starts to move with speed 2 m/s, so its kinetic energy is
$K= \frac{1}{2}m_bv_b^2$
As it compresses the spring, the speed of the block decreases and its kinetic energy is converted into elastic potential energy of the spring, which undergoes through a compression

with respect to its rest position. When the block completely stops, the compression of the spring is maximum, $x_{max}$ , and the elastic potential energy of the spring is:
$U= \frac{1}{2} k x_{max}^2$
where k=500 N/m

is the constant of the spring.
For the conservation of energy, we must have
K=U

So we can write
$\frac{1}{2}m_b v_b^2 = \frac{1}{2}kx_{max}^2$
and we can solve to find the compression of the spring:
$x_{max}= \sqrt{ \frac{m_b v_b^2}{k} }= \sqrt{ \frac{(15 kg)(2m/s)^2}{500 N/m} }=0.35 m$

Answer from: hahalol123goaway

0.35 m is the maximum distance that the block will compress the spring after the collision.

Given :

Mass, $\rm m_b = 15 \;Kg\;and\; m_s = 3\;Kg$

Spring constant, K = 500 N/m

Solution :

Before the collision the total momentum is,

$\rm p_i = m_s \times v_s$

Whare m_s is mass of stone and $\rm v_s$ is the velocity of stone beforethe collision.

After the collision stone and block both are moving therefore total momentum is,

$\rm p_f = m_sv_s' + m_bv_b$

where, $\rm v_s'$ is the velocity of stone after collision, $\rm m_b$ is the mass of block and $\rm v_b$ is the speed of the block after collision.

Now, momentum should be conserve

$\rm p_i = p_f$

$\rm m_sv_s= m_sv_s'+ m_bv_b$

$\rm v_b = \dfrac{m_s}{m_b}(v_s - v_s')$ ------- (1)

$\rm v_s = 8\;m/sec$ (Given)

$\rm v_s' = 2 \;m/sec$ (Given)

Now putting the values in equation (1) we get,

$\rm v_b = \dfrac{3}{15}(8-2)$

$\rm v_b = 2\;m/sec$

Therefore, block slides with the speed of 2 m/s to the right after the collision.

We know that the block starts to move with speed 2 m/s, so its kinetic energy is

$\rm K = \dfrac{1}{2}m_bv_b^2$

As the spring is compress, block speed decreases and its kinetic energy is converted into elastic potential energy of the spring and the elastic potential energy of the spring is,

$\rm U = \dfrac{1}{2} Kx^2_m_a_x$

Now through conservation of energy we have,

$\rm K = U$

$\rm m_bv_b^2 = Kx^2_m_a_x$

$\rm x_m_a_x = \sqrt{\dfrac{m_bv_b^2}{K}}$ ----- (2)

Putting the values of $\rm m_b,\;v_b\;and\; K$ in equation (2),

$\rm x_m_a_x= \sqrt{\dfrac{15\times2^2}{500}}$

$\rm x_m_a_x=0.35\;m$

0.35 m is the maximum distance that the block will compress the spring after the collision.

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A15 kg block is attached to a very light horizontal spring of spring constant 500 n/m and is resting...