Question

A 10.0-kg shell is traveling horizontally to the right at 25.0 m/s relative to the ground when it explodes into two fragments, one of mass 3.00 kg and the other of mass 7.00 kg. The lighter fragment goes directly forward, and the explosion releases 1.50 x 10^3 J of mechanical energy to the fragments. Find the magnitude and direction of the velocity of the heavier fragment relative to the ground just after the explosion.

Answer 1

Using the principle of conservation of momentum, we can determine the magnitude and direction of the velocity of the heavier fragment. By setting up equations for the initial and final momentum of the system, we can solve for the unknown velocity. The magnitude of the velocity of the heavier fragment relative to the ground is 35.71 m/s, and its direction is to the right.

Step-by-step explanation:

We can use the principle of conservation of momentum to solve this problem. Before the explosion, the total momentum of the system (shell + fragments) is given by:

Initial momentum = mass of shell x velocity of shell

After the explosion, the total momentum of the system is distributed between the two fragments. Let v1 be the velocity of the lighter fragment and v2 be the velocity of the heavier fragment. The total momentum can be expressed as:

Total momentum = mass of lighter fragment x velocity of lighter fragment + mass of heavier fragment x velocity of heavier fragment

Since the lighter fragment goes directly forward, its velocity is given by:

Velocity of lighter fragment = 1.50 x 10³ J / mass of lighter fragment

To find the velocity of the heavier fragment, we can substitute the known values into the equation for total momentum and solve for v2:

mass of shell x velocity of shell = mass of lighter fragment x velocity of lighter fragment + mass of heavier fragment x velocity of heavier fragment

Substituting the given values:

10.0 kg x 25.0 m/s = 3.00 kg x (1.50 x 10³ J / 3.00 kg) + 7.00 kg x velocity of heavier fragment

Simplifying the equation:

250 kg·m/s = 500 J + 7.00 kg x velocity of heavier fragment

250 kg·m/s - 500 J = 7.00 kg x velocity of heavier fragment

Solving for velocity of heavier fragment:

Velocity of heavier fragment = (250 kg·m/s - 500 J) / 7.00 kg

Calculating the velocity:

Velocity of heavier fragment = 35.71 m/s

The magnitude of the velocity of the heavier fragment relative to the ground is 35.71 m/s, and since the lighter fragment goes directly forward, the direction of the velocity of the heavier fragment relative to the ground is to the right.

Answer 2

The magnitude of the velocity of the heavier fragment relative to the ground just after the explosion is approximately 10.71 m/s, and its direction is to the left.

Step-by-step explanation:

To find the magnitude and direction of the velocity of the heavier fragment relative to the ground just after the explosion, we can use the principle of conservation of momentum. Before the explosion, the total momentum of the system is zero since the shell is traveling horizontally. After the explosion, the total momentum is still zero, so the momentum of the lighter fragment is equal in magnitude and opposite in direction to the momentum of the heavier fragment.

Using the equation for momentum: m1 * v1 + m2 * v2 = 0, where m1 and m2 are the masses of the fragments and v1 and v2 are their velocities, we can solve for v2, the velocity of the heavier fragment.

Given that the mass of the lighter fragment is 3.00 kg and its velocity is 25.0 m/s (the same as the initial velocity of the shell), and the mass of the heavier fragment is 7.00 kg, we can substitute these values into the equation to find the velocity of the heavier fragment.

Plugging in the values, we get: (3.00 kg * 25.0 m/s) + (7.00 kg * v2) = 0

Simplifying the equation, we have: 75.0 kg m/s + 7.00 kg * v2 = 0

Finally, solving for v2, we find that the magnitude of the velocity of the heavier fragment is approximately 10.71 m/s, and since it is in the opposite direction of the lighter fragment, its direction is to the left relative to the ground.