A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be

11.5 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

14.0 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.0 m/s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.89 m/s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
9.89 m/s

$P_{x} = m \times v_{x} = 1 \times 21 = 21 k g m / s$

$P_{y} = m \times v_{y} = 1 \times 21 = 21 k g m / s$

$∴$ Resultant = $\sqrt{P_{x}^{2} + P_{y}^{2}} = 21 \sqrt{2} k g m / s$

The momentum of heavier fragment should be numerically equal to resultant of ${\to P}_{x}$ and ${\to P}_{y}$ .

$3 \times v = \sqrt{P_{x}^{2} + P_{y}^{2}} = 21 \sqrt{2} ∴ v = 7 \sqrt{2} = 9.89 m / s$

Suggest Corrections

Similar questions

A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be

Q. A body of mass

18 kg

initially at rest explodes into three fragments with masses in the ratio of

2 : 1 : 3

. If smaller fragments fly in mutually perpendicular directions and their velocities are

10 m/s

and

5 m/s

respectively, then what is the speed of the heaviest fragment?

Q. A body of mass

18 kg

initially at rest explodes into three fragments with masses in the ratio of

2 : 1 : 3

. If smaller fragments fly in mutually perpendicular directions and their velocities are

10 m/s

and

5 m/s

respectively, then what is the speed of the heaviest fragment?

Q. A shell of mass

m

is at rest initially. It explodes into three fragments having masses in the ratio

2 : 2 : 1

. The fragments having equal masses fly off along mutually perpendicular direction with speed

v

. What will be the speed of the third (lighter) fragment?

Join BYJU'S Learning Program

A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be

The correct option is D 9.89 m/s Px=m×vx=1×21=21kgm/s Py=m×vy=1×21=21kgm/s ∴ Resultant = √P2x+P2y=21√2kg m/s The momentum of heavier fragment should be numerically equal to resultant of →Px and →Py. 3×v=√P2x+P2y=21√2∴v=7√2=9.89m/s