Question

A 0.5kg stone moving north at 4 m/s collides with a 4kg lump of clay moving west at 1 m/s. The stone becomes embedded in the clay. What is the velocity (magnitude and direction) of the composite body after the collision?

Answer 1

Answer:

The velocity of the composite body is 0.99m/s 63.43° west of north.

Explanation:

Here the law of conservation of energy says that

$(1).\: \: m_1v_1cos(0)+m_2v_2cos(90^o)=(m_1+m_2)v_f cos(\theta)$

$(2).\: \: m_1v_1sin(0)+m_2v_2sin(90^o)=(m_1+m_2)v_f sin(\theta)$

where $v_f$ is the final velocity if the composite body, and $\theta$ is measured from west of north.

Putting in numbers and simplifying the above equation we get:

$(3).\: \: m_1v_1=(m_1+m_2)v_f cos(\theta)$

$(4).\: \: m_2v_2=(m_1+m_2)v_f sin(\theta)$

dividing equation (4) by (3) gives

$\dfrac{ m_2v_2=(m_1+m_2)v_f sin(\theta)}{ m_1v_1=(m_1+m_2)v_f cos(\theta)}$

$\dfrac{m_2v_2}{m_1v_1} = \dfrac{sin(\theta)}{ cos(\theta)}$

$(5).\: \: tan(\theta) = \dfrac{m_2v_2}{m_1v_1}$

putting in $m_1 = 0.5kg$, $v_1 = 4m/s$, $m_2 = 4kg$, and $v_2 = 1m/s$ we get:

$tan(\theta) = \dfrac{(4kg)(1m/s)}{(0.5kg)(4m/s)}$

$\boxed{\theta = 63.43^o}$

Thus, the final velocity $v_f$ we get from equation (3) is:

$v_f=\dfrac{m_1v_1}{(m_1+m_2)cos(\theta)}$

$v_f=\dfrac{(0.5kg)(4m/s)}{(4kg+0.5kg)cos(63.43^o)}$

$\boxed{v_f = 0.99m/s}$

Thus, the velocity of the composite body is 0.99m/s 63.43° west of north.