Question

Reset
A block of mass 4.5 Kg is suspended by a string
and held by hand from a fixed support and an air
gun fires a shell of mass 0.5 kg with a velocity of 40
m/s. The shell gets embedded into the block as it
strikes it and the block is released to allow it to
move freely. What is the height attained by the block
when the tension force in the string is minimum?​

Answer 1

Answer:

conservation of energy

2

1

mv

2

=mgh

V

2

=2gh

V = 1.715 m / s

Applying

conservation of momentum

μ = ( m + M ) V

solving

U=

0.02

8.607

= 430 m / s

PLZ MARK ME AS A BRAINLIST

Answer 2

Answer:

The height attained by the block when the tension force in the string is minimum is 1.6m.

Explanation:

Since no external force is acting on the system the momentum and energy are conserved.

First, let's apply the conservation of angular momentum. Which is given as,

$mu=(m+M)v$       (1)

Where,

m=mass of the shell

M=mass of the block

u=initial velocity of the shell

v=final velocity of the embedded part

From the question we have,

m=0.5kg

M=4.5kg

u=40m/s

By substituting the values in equation (1) we get;

$0.5\times 40=(0.5+4.5)v$

$20=5\times V$

$v=4m/s$        (2)

From the conservation of energy we have,

Kinetic energy=Potential energy

$\frac{1}{2} (m+M)v^2=(m+M)gh$

$v^2=gh$         (3)

v=combined velocity of the embedded part

g=acceleration due to gravity=10m/s²

h=height attained by the block

By substituting the values in equation (3) we get;

$(4)^2=10\times h$

$h=\frac{16}{10}$

$h=1.6m$

Hence, the height attained by the block when the tension force in the string is minimum is 1.6m.

#SPJ2