Question

A shell of mass 200 gm is ejected from a gun
mass 4 kg by an explosion that generates 1.05 ki
of energy. The initial velocity of the shell is
[AIPMT (Prelims)-2008]
(1) 120 ms-1
(2) 100 ms-1
(3) 80 ms -1
(4) 40 ms -1​

Answer 1

Answer:

energy = 1/2 m v^2

1050 J = 1/2 (200/1000)v^2

v = √ 10500

approximately equal to 100

so option 2

hope it helps you

Answer 2

Given :

• m1 = 200g = 1/5 kg
• m2 = 4 kg
• Total energy ( E ) = 1.05 kJ = 1050 J

To Find :

• Initial velocity of the shell

Solution :

Let u1 be initial velocity of the shell and u2 be the initial velocity of the gun.

According to principle of conservation of linear momentum

$\tt \Large m_1u_1 + m_2u_2 = 0 \\ \\ \implies\tt u_2 = - \frac{m_1}{m_2}u_1 \\ \\ \implies\tt u_2 = \frac{ - 1}{5 \times 4} u_1 \\ \\ \implies\tt u_2 = - \frac{u_1}{20}$

Now

$\Large \tt E = \frac{1}{2} m_1 {u_1}^{2} + \frac{1}{2} m_2 {u_2}^{2} \\ \\\implies \tt1050 = \frac{1}{2} \times \frac{1}{5}{u_1}^{2} + \frac{1}{2} \times 4 \bigg( - \frac{{u_2}}{20} \bigg) ^{2} \\ \\ \tt \implies \frac{{u_1}^{2}}{10} + \frac{{u_1}^{2}}{200} = \frac{21}{200} {u_1}^{2} \\ \\ \tt \implies{u_1}^{2} = \frac{200 \times 1050}{21} \\ \\ \tt \implies {u_1}^{2} = 10000 \\ \\ \tt \implies u_1 = \sqrt{10000} \\ \\ \tt \implies u_1 = 100 \: m \: {s}^{ - 1}$

Initial velocity of shell is 100 m/s