A moving body with a mass m1 strikes a stationary body of mass m2. The masses m1 and m2 should be in the ratio m1/m2 so as to decrease the velocity of the first body 1.5 times assuming a perfectly elastic impact. Then the ratio m1/m2 is

The correct option is 5.

Let the initial velocity of the moving body of mass _m1 be _u1, and the stationary body of mass _m2 be initially at rest (_u2 = 0).

For a one-dimensional perfectly elastic collision, the final velocity _v1 of the first body is given by the formula:
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1$

According to the problem, the velocity of the first body decreases 1.5 times after the collision. This means:
$v_1=\frac{u_1}{1.5}=\frac{2}{3}{u}_1$

Equating the two expressions for _v1:
$\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1=\frac{2}{3}{u}_1$

Dividing both sides by _u1:
$\frac{m_1-m_2}{m_1+m_2}=\frac{2}{3}$

Cross-multiplying to solve for the ratio:
$3(m_1-m_2)=2(m_1+m_2)$
$3m_1-3m_2=2m_1+2m_2$
$m_1=5m_2$

Thus, the ratio of the masses is:
$\frac{m_1}{m_2}=5$