Question

In midair an M = 135 kg bomb explodes into two pieces of m1 = 105 kg and another, respectively. Before the explosion, the bomb was moving at 20.0 m/s to the east. After the explosion, the velocity of the m1 = 105 kg piece is 62.0 m/s to the east. Find the velocity (in m/s) (with a proper sign) of the other piece after the explosion.

Answer 1

Answer:

-117 m/s

Explanation:

We are given that

Mass, M=135 kg

$m_1=105 kg$

Let m be the mass of another piece

Mass of another piece=135-105=30 kg

V=20 m/s

$v_1=62 m/s$

We have to find the velocity of the other piece.

According to law of conservation of momentum

$MV=m_1v_1+m_2v_2$

Substitute the values

$150\times 20=105\times 62+30v_2$

$3000=6510+30v_2$

$3000-6510=30v_2$

$3510=30v_2$

$v_2=\frac{-3510}{30}=-117m/s$

Hence, the velocity of other piece after the explosion=-117 m/s

Answer 2

Answer:

Explanation:

M = 135 kg

U = 20 m/s East

m1 = 105 kg

m2 = M - m1 = 135 - 105 = 30 kg

v1 = 62 m/s East

let the velocity of another part is v2.

Use conservation of momentum

M x U = m1 x v1 + m2 x v2

135 x 20 = 105 x 62 + 30 x v2

2700 = 6510 + 30 v2

v2 = - 127 m/s

Thus, the velocity of another part is 127 m/s due west.