Question

in the figure shown the block a collides head on with another block B at rest. mass of B is twice the mass of the A.the block A stops after collision the coefficient of restitution is​

Answer 1

Answer:

Coefficient of restitution (e) = 0.5

Explanation:

Given,

• Mass of the block A = m
• Mass of the block B = 2m
• Final  velocity of the block A = $v_1\ =\ 0$
• Initial velocity of the block B = $u_2\ =\ 0$

Let v be v_f be the the initial speed of the block A and final speed of the block B after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ m_1v_1\ +\ m_2v_2\\\Rightarrow mv\ +\ 2m\times 0\ =\ m\times 0\ +\ 2mv_f\\\Rightarrow v\ =\ 2v_f\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Now from the equation of restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}$

Now substituting the values in the above expression of restitution, we get,

$\Rightarrow e\ =\ \dfrac{0\ -\ v_f}{0\ -\ v}\\\Rightarrow e\ =\ \dfrac{v_f}{v}\\\Rightarrow e\ =\ \dfrac{v_f}{2v_f}\\\Rightarrow e\ =\ 0.5$

Answer 2

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