A particle of mass 3 m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2 m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to

Question

A particle of mass 3 m is projected from the ground at some angle with horizontal. The horizontal range is R. At the highest point of its path it breaks into two pieces m and 2 m. The smaller mass comes to rest and larger mass finally falls at a distance x from the point of projection where x is equal to (A) (3 R /4) (B) (3 R /2) (C) (5 R /4) (D) 3 R

Tardigrade Admin · Accepted Answer

x cm =( m 1 x 1+ m 2 x 2/ m 1+ m 2) R =( m (( R /2))+2 m ( x 2)/3 m ) ⇒ x 2=(5 R /4) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1102a35d0e28a30e5c5e2f1c0203ca7d-.png" />

$\frac{3 R}{4}$

$\frac{3 R}{2}$

$\frac{5 R}{4}$

$3 R$

$x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}$
$R = \frac{m ( \frac{R}{2} ) + 2 m ( x _{2} )}{3 m}$
$\Rightarrow x_{2} = \frac{5 R}{4}$

43R​

23R​

45R​

3R

xcm​=m1​+m2​m1​x1​+m2​x2​​ R=3mm(2R​)+2m(x2​)​ ⇒x2​=45R​

$\frac{3 R}{4}$

$\frac{3 R}{2}$

$\frac{5 R}{4}$

$3 R$

$x_{c m} = \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}}$
$R = \frac{m ( \frac{R}{2} ) + 2 m ( x _{2} )}{3 m}$
$\Rightarrow x_{2} = \frac{5 R}{4}$