A glass ball, ball A, of mass 6.0 g moves at a velocity of 19.0 cm/s. It collides with a second ball, ball B, of mass 9.0 g, moving along the same line of velocity of 11.0 cm/s. After the collison, ball A is still moving, but with a velocity of 8.0 cm/s. a) What was ball A's original momentum? What is ball A's change in momentum? c) What is ball B's change in momentum? d) What is the momentum of ball B after the collison? e) What is ball B's speed after the collison?
Linear momentum is conserved. Whatever the momentum A lost in the collision, B gained.
I will be happy to critique your thinking.
Ball A's original momentum is .00114 kgm/s. The change in Ball A's momentum would be .006 kg x (.19 - .08) m/s, correct? Ball B's change in momentum would be .009 kg x (.11 - v) m/s. The momentum of Ball B would be .009 x the change in velocity from the previous question. Ball B's speed would be the original velocity minus the change in velocity.
Ball A's original momentum is .00114 kgm/s. The change in Ball A's momentum would be .006 kg x (.19 - .08) m/s, correct? Ball B's change in momentum would be .009 kg x (.11 - v) m/s. The momentum of Ball B would be .009 x the change in velocity from the previous question. Ball B's speed would be the original velocity minus the change in velocity.
Correct on the change of A momentum.
Ball B change in momentum is equal to the change in A momentum, not the change in A velocity.
New Ball B momentum:
.009*.11+ .006(.19-.08)
From that, you determine B velocity (by dividing the Momentum my mass B)
To answer these questions, let's use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
a) First, let's calculate the original momentum of ball A. The momentum (p) of an object is given by the product of its mass (m) and velocity (v). So, the original momentum of ball A is:
p(A) = m(A) * v(A)
p(A) = 6.0 g * 19.0 cm/s
To simplify calculations, let's convert the mass of ball A to kilograms and the velocity to meters per second:
p(A) = 0.006 kg * 0.19 m/s
p(A) = 0.00114 kg·m/s
Therefore, the original momentum of ball A is 0.00114 kg·m/s.
b) The change in momentum (Δp) of an object is equal to its final momentum (p_f) minus its initial momentum (p_i). So, the change in momentum of ball A is:
Δp(A) = p_f(A) - p_i(A)
Δp(A) = 0.006 kg * 0.08 m/s - 0.006 kg * 0.19 m/s
Let's calculate the change in momentum of ball A:
Δp(A) = 0.00048 kg·m/s - 0.00114 kg·m/s
Δp(A) = -0.00066 kg·m/s
Therefore, the change in momentum of ball A is -0.00066 kg·m/s.
c) Similarly, let's calculate the change in momentum of ball B. The mass and velocity values given for ball B are its initial values. The change in momentum of ball B can be calculated using the same formula:
Δp(B) = p_f(B) - p_i(B)
Δp(B) = m(B) * v_f(B) - m(B) * v_i(B)
Δp(B) = 9.0 g * 0.0 cm/s - 9.0 g * 11.0 cm/s
Converting to SI units:
Δp(B) = 0.009 kg * 0.0 m/s - 0.009 kg * 0.11 m/s
Δp(B) = -0.00099 kg·m/s
Therefore, the change in momentum of ball B is -0.00099 kg·m/s.
d) Now, let's calculate the momentum of ball B after the collision. Since the total momentum is conserved, we can find the momentum of ball B by adjusting for the change in momentum:
p(B)_f = Δp(B) + p(B)_i
p(B)_f = -0.00099 kg·m/s + 0.009 kg * 0.11 m/s
Calculating:
p(B)_f = -0.00099 kg·m/s + 0.00099 kg·m/s
p(B)_f = 0.000 kg·m/s
Therefore, the momentum of ball B after the collision is 0.000 kg·m/s.
e) Finally, let's determine the speed of ball B after the collision. Since speed is the magnitude of velocity, we can use the given information about the velocity to calculate it:
speed(B)_f = |v(B)_f|
speed(B)_f = |0.0 cm/s|
Converting to SI units:
speed(B)_f = |0.0 m/s|
speed(B)_f = 0.0 m/s
Therefore, the speed of ball B after the collision is 0.0 m/s.