a bolwing ball of mass 2.00 kg strikes a stationary pin of mass 5.00 x 10^2 g. the collision lasts for .60s after which the pin moves off with a velocity of 12.0 m/s [w]
a)accel of pin during the collision
b)force exerted by bowling ball on the pin
c)the accel of the bowling ball during collision
a) acceleration = (Pin velocity change)/(0.60 s)
= 20 m/s^2 = a
b) F = M(pin)*a = 0.500 kg*20 m/s^2
= 10 Newtons
The forces F on ball and pin are equal and opposite, so
c) a(ball) = F/M(ball) = 10/2 = 5 m/s^2
a) The acceleration of the pin during the collision can be calculated using the equation:
Acceleration = (Change in velocity) / (Time taken)
Since the pin was initially stationary and ends up moving with a velocity of 12.0 m/s, the change in velocity is 12.0 m/s. The time taken is 0.60 s.
Acceleration = 12.0 m/s / 0.60 s = 20.0 m/s²
b) The force exerted by the bowling ball on the pin can be calculated using Newton's second law of motion:
Force = mass × acceleration
The mass of the pin is 5.00 x 10^2 g, which is equivalent to 0.500 kg. The acceleration of the pin during the collision is 20.0 m/s².
Force = 0.500 kg × 20.0 m/s² = 10.0 N
c) The acceleration of the bowling ball during the collision can be calculated using the equation:
Acceleration = Force / mass
The mass of the bowling ball is 2.00 kg, and the force exerted by the bowling ball on the pin is 10.0 N.
Acceleration = 10.0 N / 2.00 kg = 5.0 m/s²
So, the acceleration of the bowling ball during the collision is 5.0 m/s².
To solve this problem, we can use the principles of physics related to momentum and force. Let's break down each part of the problem step by step:
a) Acceleration of the pin during the collision:
To find the acceleration of the pin during the collision, we can use the formula for acceleration:
acceleration = (final velocity - initial velocity) / time
Given:
Mass of the pin (m1) = 5.00 x 10^2 g = 0.5 kg
Final velocity of the pin (v1f) = 12.0 m/s
Initial velocity of the pin (v1i) = 0 (since it is stationary)
Time (t) = 0.60 s
Plugging in the values into the formula, we have:
acceleration = (v1f - v1i) / t
acceleration = (12.0 m/s - 0) / 0.60 s
acceleration = 20.0 m/s^2
Therefore, the acceleration of the pin during the collision is 20.0 m/s^2.
b) Force exerted by the bowling ball on the pin:
To find the force exerted by the bowling ball on the pin, we can use the formula for force:
force = mass x acceleration
Given:
Mass of the bowling ball (m2) = 2.00 kg
Acceleration of the pin (a) = 20.0 m/s^2
Plugging in the values into the formula, we have:
force = m2 x a
force = 2.00 kg x 20.0 m/s^2
force = 40.0 N
Therefore, the force exerted by the bowling ball on the pin is 40.0 N.
c) Acceleration of the bowling ball during the collision:
To find the acceleration of the bowling ball during the collision, we can use the formula for acceleration:
acceleration = (final velocity - initial velocity) / time
Given:
Mass of the bowling ball (m2) = 2.00 kg
Final velocity of the bowling ball (v2f) = ?
Initial velocity of the bowling ball (v2i) = 0 (since it strikes a stationary pin)
Time (t) = 0.60 s
We need to rearrange the formula to solve for the final velocity:
final velocity = (acceleration x time) + initial velocity
v2f = (acceleration x t) + v2i
Plugging in the values into the formula, we have:
v2f = (acceleration x t) + v2i
v2f = (20.0 m/s^2 x 0.60 s) + 0
v2f = 12.0 m/s
Therefore, the acceleration of the bowling ball during the collision is 12.0 m/s.
To solve this problem, we can use the principles of conservation of momentum and Newton's second law of motion. Let's break it down step by step:
a) Acceleration of the pin during the collision:
We can find the acceleration of the pin by using the equation of motion:
v = u + at,
where:
v = final velocity of the pin (12.0 m/s)
u = initial velocity of the pin (0 m/s)
a = acceleration of the pin during the collision (to be determined)
t = time for which the collision lasts (0.60 s)
Substituting the values into the equation, we have:
12.0 m/s = 0 m/s + a * 0.60 s.
Simplifying the equation, we can solve for 'a':
a = (12.0 m/s - 0 m/s) / 0.60 s.
a = 20 m/s^2.
Therefore, the acceleration of the pin during the collision is 20 m/s^2.
b) Force exerted by the bowling ball on the pin:
We can calculate the force exerted by the bowling ball on the pin using Newton's second law of motion:
F = m * a,
where:
F = force exerted by the bowling ball on the pin (to be determined)
m = mass of the pin (5.00 x 10^2 g converted to kilograms, which is 0.500 kg)
a = acceleration of the pin during the collision (20 m/s^2)
Substituting the values into the equation, we have:
F = 0.500 kg * 20 m/s^2.
F = 10 N.
Therefore, the force exerted by the bowling ball on the pin is 10 Newtons.
c) Acceleration of the bowling ball during the collision:
Since the pin is stationary initially, its velocity is zero. The bowling ball moves towards the pin and hits it, causing the pin to move off with a velocity of 12.0 m/s.
By applying the conservation of momentum principle, the momentum before the collision is equal to the momentum after the collision.
Momentum before the collision = Momentum after the collision.
(mass of bowling ball) * (initial velocity of the bowling ball) = (mass of pin) * (final velocity of the pin).
(2.00 kg) * (initial velocity of the bowling ball) = (0.500 kg) * (12.0 m/s).
Simplifying the equation, we can solve for the initial velocity of the bowling ball:
(initial velocity of the bowling ball) = (0.500 kg * 12.0 m/s) / 2.00 kg.
(initial velocity of the bowling ball) = 3.00 m/s.
As the duration of the collision is given to be 0.60 s, we can use the equation of motion to find the acceleration of the bowling ball:
v = u + at,
where:
v = final velocity of the bowling ball (0 m/s)
u = initial velocity of the bowling ball (3.00 m/s)
a = acceleration of the bowling ball during the collision (to be determined)
t = time for which the collision lasts (0.60 s)
Substituting the values into the equation, we have:
0 m/s = 3.00 m/s + a * 0.60 s.
Simplifying the equation, we can solve for 'a':
a = (0 m/s - 3.00 m/s) / 0.60 s.
a = -5.0 m/s^2.
Therefore, the acceleration of the bowling ball during the collision is -5.0 m/s^2.