Find the final velocity of mass A?

Question

Two gliders, A(.55kg) and B(.35kg) are held together by a string which compresses a spring between the gliders. The masses are traveling at 0.5m/s. a match burns the string and mass B is seen to be traveling at 2.1 m/s. Find the final velocity of mass A.

Answer 1

Conservation of momentum states that the momentum of a group of objects is the same before and after a collision/explosion. Initial momentum = final momentum, p_i = p_f .

The initial momentum of the two gliders before the explosion, is equal to the final momentum of gliders A and B added together.

GIVEN:

m_A = 0.55kg

m_{B ⁼} 0.35kg

v_i = +0.5m/s

v_f of m_B = +2.1 m/s

 

UNKNOWN:

v_f of m_A = ?

 

EQUATION:

p = mv

p_i = p_f

(mv)_i = (mv)_f  

(m_A + m_B)v_i = (m_Av_Af) + (m_Bv_Bf)

The initial momentum is that one one object since the gliders are tied together. The final momentum is the sum of the momenta of glider A and glider B.

 

SOLUTION:

(m_A + m_B)v_i = (m_A v_Af) + (m_B v_Bf) substitute the variables with their values and solve for v_Af , the final velocity of glider A.

 

(0.55 kg + 0.35 kg)0.5 m/s = (0.55 kg • v_Af) + (0.35 kg • 2.1 m/s)

0.45 kg•m/s = (0.55 kg • v_Af) + (0.735 kg•m/s)

(0.45 kg•m/s - 0.735 kg•m/s) = (0.55kg • v_Af)

(-0.285 kg•m/s) / (0.55 kg) = v_Af 

-0.52 m/s = v_Af

After the explosion, glider A moves off at -0.52 m/s. The negative sign indicates it moves in the opposite direction of glider B.