Question

An object of mass $m$ sliding along a frictionless surface collides with the stationary object of mass $m$. The two bodies stick together. If the kinetic energy of the two-body system is $E$. Calculate the initial velocity of the first object before impact?

• A. $\sqrt{E/2m}$
• B. $\sqrt{2E/2m}$
• C. $\sqrt{2E/m}$
• D. $\sqrt{E/m}$
• E. $2\sqrt{E/m}$

Answer 1

The correct option is E $2\sqrt{E/m}$

As bodies stick together after the collision so the collision is inelastic , but we know that linear momentum remains conserved also in an inelastic collision,

Momentum before collision = Momentum after collision

$m{v}_1+0=(m+m)v$ where $v_1$ is the velocity of first object and given that, a body is at rest before collision so its momentum is taken zero,

or $v=v_1/2$

Now the kinetic energy of two body system

$E=\frac{1}{2}(m+m)v^2$

or $E=\frac{1}{2}2m{v}^2$

or $E=m{v}^2$

or $E=m{\left(\frac{v_1}{2}\right)}^2$

or $v_1=\sqrt{4E/m}=2\sqrt{E/m}$