The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

Question

The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/08496f3670c6fc5ff82352bb64e12424-.png" /> (A) Zero (B) (M L2/K) (C) √M K L (D) (K L2/2 M)

Tardigrade Admin · Accepted Answer

When collision occurs all kinetic energy it possesses get converted into P.E (1/2) M v2=(1/2) K L2 ⇒ M v2=K L2 Multiplying both side by M. (Mass) (M v)2=M K L2 M v=√M K L2 P=√M K L or √M K L2

Q. The block of mass $M$ moving on the frictionless horizontal surface collides with the spring of spring constant $K$ and compresses it by length $L$ . The maximum momentum of the block after collision is

Zero

$\frac{M L ^{2}}{K}$

$M K L$

$\frac{K L ^{2}}{2 M}$

When collision occurs all kinetic energy it possesses get converted into P.E
$\frac{1}{2} M v^{2} = \frac{1}{2} K L^{2}$
$\Rightarrow M v^{2} = K L^{2}$
Multiplying both side by M. (Mass)
$(M v)^{2} = M K L^{2}$
$M v = M K L^{2}$
$P = M K L$ or $M K L^{2}$

Q. The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is

Zero

KML2​

MK​L

2MKL2​

When collision occurs all kinetic energy it possesses get converted into P.E 21​Mv2=21​KL2 ⇒Mv2=KL2 Multiplying both side by M. (Mass) (Mv)2=MKL2 Mv=MKL2​ P=MK​L or MKL2​

Q. The block of mass $M$ moving on the frictionless horizontal surface collides with the spring of spring constant $K$ and compresses it by length $L$ . The maximum momentum of the block after collision is

$\frac{M L ^{2}}{K}$

$M K L$

$\frac{K L ^{2}}{2 M}$

When collision occurs all kinetic energy it possesses get converted into P.E
$\frac{1}{2} M v^{2} = \frac{1}{2} K L^{2}$
$\Rightarrow M v^{2} = K L^{2}$
Multiplying both side by M. (Mass)
$(M v)^{2} = M K L^{2}$
$M v = M K L^{2}$
$P = M K L$ or $M K L^{2}$