160k views
0 votes
A bomb explodes into three pieces A, B, and C of equal mass. Piece A flies with a speed of 40.0 m/s, and piece B with a speed of 30.0 m/s at an angle of 90° relative to the direction of A as shown in the Figure. Determine the speed of piece C and the direction of its velocity relative to the direction of piece A.

1 Answer

5 votes

Answer:

Step-by-step explanation:

Let's use conservation of momentum to solve this problem. Since the system was initially at rest, the total momentum of the system before the explosion is zero. After the explosion, the total momentum of the system is still zero since there are no external forces acting on the system. Therefore, the momentum of each piece must add up to zero.Let the mass of each piece be m. Then the velocity of piece A is 40.0 m/s to the right and the velocity of piece B is 30.0 m/s upward. Let the velocity of piece C be v with an angle θ relative to the direction of piece A.Using conservation of momentum, we have:(m)(40.0 m/s) + (m)(30.0 m/s) + (m)(v cos θ) = 0Simplifying and solving for v, we get:v cos θ = -(70.0 m/s)Now, using conservation of kinetic energy, we know that the total kinetic energy of the system after the explosion is equal to the kinetic energy before the explosion, since there are no external forces acting on the system. The kinetic energy of each piece is given by:KE = (1/2)mv^2Therefore, we have:(1/2)m(40.0 m/s)^2 + (1/2)m(30.0 m/s)^2 + (1/2)m(v^2) = (1/2)m(70.0 m/s)^2Simplifying and solving for v, we get:v = ±50.0 m/sSince piece C must move in the opposite direction to pieces A and B in order to cancel out their momenta, we take the negative value of v. Therefore, the speed of piece C is 50.0 m/s and its direction is 180° (opposite) relative to the direction of piece A.

User Yoyojs
by
7.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.