A particle of mass 80 units is moving with a uniform speed v=4√2 unit in XY. plane, along a fine y = x + 5. The magnitude of the angular momentum of the particle about the origin is.

Question

A particle of mass 80 units is moving with a uniform speed v=4√2 unit in XY. plane, along a fine y = x + 5. The magnitude of the angular momentum of the particle about the origin is. (A) 1600 units (B) 160√2 unit (C) 152√2 (D) 16√2

Tardigrade Admin · Accepted Answer

barL = barr × barp = m ( barr × barv)

Q. A particle of mass 80 units is moving with a uniform speed $v = 42$ unit in XY. plane, along a fine y = x + 5. The magnitude of the angular momentum of the particle about the origin is.

1600 units

$1602$ unit

$1522$

$162$

$\overset{ˉ}{L} = \overset{r}{ˉ} \times \overset{p}{ˉ} = m (\overset{r}{ˉ} \times \overset{v}{ˉ})$

Q. A particle of mass 80 units is moving with a uniform speed v=42​ unit in XY. plane, along a fine y = x + 5. The magnitude of the angular momentum of the particle about the origin is.

1600 units

1602​ unit

1522​

162​

Lˉ=rˉ×pˉ​=m(rˉ×vˉ)

Q. A particle of mass 80 units is moving with a uniform speed $v = 42$ unit in XY. plane, along a fine y = x + 5. The magnitude of the angular momentum of the particle about the origin is.

$1602$ unit

$1522$

$162$

$\overset{ˉ}{L} = \overset{r}{ˉ} \times \overset{p}{ˉ} = m (\overset{r}{ˉ} \times \overset{v}{ˉ})$