A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. Assuming coefficient of friction between the skater and the ice be μ , what will be the distance across which the skater will slide back ?

Question

A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. Assuming coefficient of friction between the skater and the ice be μ , what will be the distance across which the skater will slide back ? (A) (M2 v2/2 m μ g) (B) (Mv2/2 m2 μ g) (C) (M2 v2/2 m2 μ g) (D) (M2 v2/2 m2 μ 2 g)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-cicx7xvudmoi30hm.jpg" /> Pi=0 ...........(i) Pf=MV-mV1 ..............(ii) MV-mV1=0⇒ v1=(M/m)V using 02=v12-2ax ⇒ v12=2μ gx ⇒ ((M V/m))2=2μ gx.∴ x=(M2 V2/2 m2 μ g)

Q. A skater of mass $m$ standing on ice throws a stone of mass $M$ with a velocity of $v$ in a horizontal direction. Assuming coefficient of friction between the skater and the ice be $μ$ , what will be the distance across which the skater will slide back ?

$\frac{M ^{2} v ^{2}}{2 m μg}$

$\frac{M v ^{2}}{2 m ^{2} μg}$

$\frac{M ^{2} v ^{2}}{2 m ^{2} μg}$

$\frac{M ^{2} v ^{2}}{2 m ^{2} μ ^{2} g}$

$P_{i} = 0$ ...........(i)
$P_{f} = M V - m V_{1}$ ..............(ii)
$M V - m V_{1} = 0 \Rightarrow v_{1} = \frac{M}{m} V$
using $0^{2} = v_{1}^{2} - 2 a x$
$\Rightarrow v_{1}^{2} = 2 μgx$
$\Rightarrow (\frac{M V}{m})^{2} = 2 μgx . ∴ x = \frac{M ^{2} V ^{2}}{2 m ^{2} μg}$

Q. A skater of mass m standing on ice throws a stone of mass M with a velocity of v in a horizontal direction. Assuming coefficient of friction between the skater and the ice be μ , what will be the distance across which the skater will slide back ?

2mμgM2v2​

2m2μgMv2​

2m2μgM2v2​

2m2μ2gM2v2​