A Bullet of a mass 10g is shot into a water melon of a mass 0.2 kg which is resting on a platform. At the time of impact, the bullet is travelling horizontally at 20 m/s. calculate the common velocity just after impact
momentum is conserved
m*20=(m+M)V
solve for V
To calculate the common velocity just after impact, we can begin by applying the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, provided that no external forces are acting on the system.
In this case, we have a bullet and a watermelon colliding with each other. Let's label the velocity of the bullet before the collision as v1, the mass of the bullet as m1, the velocity of the watermelon before the collision as v2, and the mass of the watermelon as m2.
According to the problem, the bullet has a mass of 10g, which is 0.01 kg, and it is traveling horizontally at a velocity of 20 m/s. The watermelon has a mass of 0.2 kg and is initially at rest.
Before the collision, the total momentum is given by:
P_initial = m1 * v1 + m2 * v2
Since the watermelon is initially at rest, v2 is equal to 0:
P_initial = m1 * v1 + m2 * 0
The bullet's momentum before the collision is given by:
P_bullet_initial = m1 * v1
Substituting the values, we get:
P_bullet_initial = 0.01 kg * 20 m/s = 0.2 kg·m/s
Since momentum is conserved, the total momentum after the collision should also be 0.2 kg·m/s.
Let the common velocity just after impact be v_common.
The total momentum after the collision is given by:
P_final = (m1 + m2) * v_common
Substituting the values and equating it to the initial momentum, we get:
P_final = (0.01 kg + 0.2 kg) * v_common = 0.2 kg·m/s
Simplifying the equation, we can solve for v_common:
0.21 kg * v_common = 0.2 kg·m/s
v_common = 0.2 kg·m/s / 0.21 kg
Calculating this, we find that v_common is approximately 0.952 m/s.
Therefore, the common velocity just after impact is approximately 0.952 m/s.