Two bowling balls of different masses collide while accelerating at 1.875 m/s^2. The first ball, rolling to the right, weighs 7 kg. The second ball, rolling to the left, weighs 4 kg. What force does the bowling ball weighing 7 kg exert on the bowling ball weighing 4 kg?

To find the force that the 7 kg bowling ball exerts on the 4 kg bowling ball, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration (F = m * a). In this case, we need to consider the scenario just before the collision.

**Step 1: Determine the forces acting on each ball due to their acceleration.**

Both bowling balls are accelerating:

- For the 7 kg bowling ball (to the right):
- Force \( F_1 = m_1 \cdot a = 7 \, \text{kg} \cdot 1.875 \, \text{m/s}^2 = 13.125 \, \text{N} \, \text{(to the right)} \)

- For the 4 kg bowling ball (to the left):
- Force \( F_2 = m_2 \cdot a = 4 \, \text{kg} \cdot 1.875 \, \text{m/s}^2 = 7.5 \, \text{N} \, \text{(to the left)} \)

**Step 2: Analyze the forces during the collision.**

When the two balls collide, we consider that they exert forces on each other based on their masses and the accelerations they've been experiencing. The force exerted by one ball on the other can also be understood in terms of Newton's third law of motion, which states that for every action, there is an equal and opposite reaction.

Since the balls are colliding, considering the motion just before the collision:

- The 7 kg bowling ball will exert a force on the 4 kg bowling ball in the direction of its motion (to the right).
- The force that the 7 kg bowling ball exerts on the 4 kg bowling ball due to its acceleration can be simplified through the concept of effective acceleration during collision.

**Step 3: Use Newton's third law.**

According to Newton's third law, the force exerted by the 7 kg bowling ball on the 4 kg bowling ball will be equal to the force exerted by the 4 kg bowling ball on the 7 kg bowling ball in magnitude but opposite in direction.

So the force exerted by the 7 kg bowling ball on the 4 kg bowling ball is equal to the contact force just before their respective accelerations (as they are colliding):

Using the calculations from Step 1, focusing now on the context of the collision rather than the general acceleration of the systems.

Thus:
1. The required answer for the force exerted by the 7 kg ball on the 4 kg ball during collision can be simply equated and understood based upon the equal and opposite reactions relative to interaction forces.
2. Therefore, within this specific analysis of the scenario of an elastic collision, the force exerted \( F \) can also be concluded as equal and such as:
\[ F = F_1 = (7 \, \text{kg} \cdot 1.875 \, \text{m/s}^2) \, or \, F = 13.125 \, \text{N} \, \text{to the right}\]

Hence the force exerted by the 7 kg bowling ball on the 4 kg bowling ball during their interaction is also \( 13.125 \, \text{N} \, \) in the opposite directional sense of the second ball’s momentum.

Therefore,
**Force exerted by the 7 kg bowling ball = 13.125 N (to the right)**.