Question

A body is traveling at 5.07 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion separates the body into two parts, each of 3.90 kg, and increases the total kinetic energy by 100 J. The forward part continues to move in the original direction of motion. What are the speeds of (a) the rear part and (b) the forward part

Answer 1

Answer:

(a). Rear part has a velocity $v_{2}$ = 0

(b). Forward part has a velocity $v_{1}$ = 4 $\frac{m}{s}$

Explanation:

Given that

$V_{0}$ = 5.07 $\frac{m}{s}$

mass m = 3.9 kg

ΔK = 100 J

(a). From conservation of momentum principal

$m v_{0} = m_{1} v_{1} + m_{2} v_{}$

Since $v_{2} = 4 - v_{1}$

the kinetic energy

ΔK = $\frac{1}{2} m (v_{1} ^{2} + v_{2} ^{2} - 2 v_{0} ^{2} )$

By substitution we get

$v_{2} ^{2} = 16 - v_{1} ^{2}$

From the momentum equation we get

$(4 - v_{1}^{2} ) = (16 -v_{1}^{2} )$

By solving the above equation we get two solutions.

$v_{1} = 0 , v_{2} = 4 \frac{m}{s}$

$v_{1} = 4 \ \frac{m}{s} , v_{2} = 0$

Since forward part is moving  in the original direction of motion so speed of the rear part is zero.

(a). Rear part has a velocity $v_{2}$ = 0

(b). Forward part has a velocity $v_{1}$ = 4 $\frac{m}{s}$